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Can a planet have a different gravitational pull depending on its location in orbit around its sun?


How/would the distance from a planet to its star affect the strength of its gravitational pull?How big can a moon be where you can physically jump out of its orbit, to its planet?Can a star orbit around a planet?What major event could disrupt planet Earth's orbit around the Sun?Could a habitable tidally locked planet have a day and night cycle caused by the eccentricity of its orbit?Can a dual planet have an orbit around the Sun similar to Earth's?Can one or multiple moon(s) pull the sea around my planet?What are the realistic problems of a planet orbiting too close to its sun?Can a Large Planet Orbit a Smaller Planet?The effects of gravitational pull in a solar system with a pendulum Sun













2












$begingroup$


As the title say, can a planet have a different gravitational pull depending on its location in orbit around its sun?



Is this explainable without magic?










share|improve this question









New contributor




Jae is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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  • $begingroup$
    Are you asking for the planet's own pull on objects that are not the Sun to vary, rather than the Sun's pull on the planet or vice versa?
    $endgroup$
    – Spencer
    6 hours ago










  • $begingroup$
    Well you could explain it with "sufficiently advanced technology", if you prefer. But no; most of the gravity you feel will be a factor of the mass beneath your feet, and that mass is not going to change dramatically during the year without magic, "magic" or civilisation-endingly apocalyptic events.
    $endgroup$
    – Starfish Prime
    6 hours ago










  • $begingroup$
    With an elliptical orbit, you can obviously be closer to the sun at some points than at other points, and one side of the planet will then have the sun + earths gravity, and the other, the earths gravity - suns, but this difference is likely to be small.
    $endgroup$
    – Tyler S. Loeper
    2 hours ago
















2












$begingroup$


As the title say, can a planet have a different gravitational pull depending on its location in orbit around its sun?



Is this explainable without magic?










share|improve this question









New contributor




Jae is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$











  • $begingroup$
    Are you asking for the planet's own pull on objects that are not the Sun to vary, rather than the Sun's pull on the planet or vice versa?
    $endgroup$
    – Spencer
    6 hours ago










  • $begingroup$
    Well you could explain it with "sufficiently advanced technology", if you prefer. But no; most of the gravity you feel will be a factor of the mass beneath your feet, and that mass is not going to change dramatically during the year without magic, "magic" or civilisation-endingly apocalyptic events.
    $endgroup$
    – Starfish Prime
    6 hours ago










  • $begingroup$
    With an elliptical orbit, you can obviously be closer to the sun at some points than at other points, and one side of the planet will then have the sun + earths gravity, and the other, the earths gravity - suns, but this difference is likely to be small.
    $endgroup$
    – Tyler S. Loeper
    2 hours ago














2












2








2





$begingroup$


As the title say, can a planet have a different gravitational pull depending on its location in orbit around its sun?



Is this explainable without magic?










share|improve this question









New contributor




Jae is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




As the title say, can a planet have a different gravitational pull depending on its location in orbit around its sun?



Is this explainable without magic?







planets science-fiction gravity solar-system rogue-planets






share|improve this question









New contributor




Jae is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|improve this question









New contributor




Jae is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|improve this question




share|improve this question








edited 5 hours ago









L.Dutch

90.6k29210437




90.6k29210437






New contributor




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asked 6 hours ago









JaeJae

112




112




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New contributor





Jae is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Jae is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











  • $begingroup$
    Are you asking for the planet's own pull on objects that are not the Sun to vary, rather than the Sun's pull on the planet or vice versa?
    $endgroup$
    – Spencer
    6 hours ago










  • $begingroup$
    Well you could explain it with "sufficiently advanced technology", if you prefer. But no; most of the gravity you feel will be a factor of the mass beneath your feet, and that mass is not going to change dramatically during the year without magic, "magic" or civilisation-endingly apocalyptic events.
    $endgroup$
    – Starfish Prime
    6 hours ago










  • $begingroup$
    With an elliptical orbit, you can obviously be closer to the sun at some points than at other points, and one side of the planet will then have the sun + earths gravity, and the other, the earths gravity - suns, but this difference is likely to be small.
    $endgroup$
    – Tyler S. Loeper
    2 hours ago

















  • $begingroup$
    Are you asking for the planet's own pull on objects that are not the Sun to vary, rather than the Sun's pull on the planet or vice versa?
    $endgroup$
    – Spencer
    6 hours ago










  • $begingroup$
    Well you could explain it with "sufficiently advanced technology", if you prefer. But no; most of the gravity you feel will be a factor of the mass beneath your feet, and that mass is not going to change dramatically during the year without magic, "magic" or civilisation-endingly apocalyptic events.
    $endgroup$
    – Starfish Prime
    6 hours ago










  • $begingroup$
    With an elliptical orbit, you can obviously be closer to the sun at some points than at other points, and one side of the planet will then have the sun + earths gravity, and the other, the earths gravity - suns, but this difference is likely to be small.
    $endgroup$
    – Tyler S. Loeper
    2 hours ago
















$begingroup$
Are you asking for the planet's own pull on objects that are not the Sun to vary, rather than the Sun's pull on the planet or vice versa?
$endgroup$
– Spencer
6 hours ago




$begingroup$
Are you asking for the planet's own pull on objects that are not the Sun to vary, rather than the Sun's pull on the planet or vice versa?
$endgroup$
– Spencer
6 hours ago












$begingroup$
Well you could explain it with "sufficiently advanced technology", if you prefer. But no; most of the gravity you feel will be a factor of the mass beneath your feet, and that mass is not going to change dramatically during the year without magic, "magic" or civilisation-endingly apocalyptic events.
$endgroup$
– Starfish Prime
6 hours ago




$begingroup$
Well you could explain it with "sufficiently advanced technology", if you prefer. But no; most of the gravity you feel will be a factor of the mass beneath your feet, and that mass is not going to change dramatically during the year without magic, "magic" or civilisation-endingly apocalyptic events.
$endgroup$
– Starfish Prime
6 hours ago












$begingroup$
With an elliptical orbit, you can obviously be closer to the sun at some points than at other points, and one side of the planet will then have the sun + earths gravity, and the other, the earths gravity - suns, but this difference is likely to be small.
$endgroup$
– Tyler S. Loeper
2 hours ago





$begingroup$
With an elliptical orbit, you can obviously be closer to the sun at some points than at other points, and one side of the planet will then have the sun + earths gravity, and the other, the earths gravity - suns, but this difference is likely to be small.
$endgroup$
– Tyler S. Loeper
2 hours ago











4 Answers
4






active

oldest

votes


















5












$begingroup$

It already happens.



Look at the schematic drawing below, where you see a planet and its Sun (not to scale), with two different places on the planet: one facing the Sun, the other on the opposite side from it.



schematic of the gravitational set up



Gravitational pull can be schematized with a vector, and the resulting pull is the result of the summation of all the pulls in a given spot.



In the place b in the picture, the pulls of the Sun and of the planet will operate in the same direction, while in a the pull from the Sun will partially counter the pull from the planet.



Therefore in b the pull will be stronger than in a.



Also, during the course of an orbit around the Sun, the distance between the two bodies changes slightly, and as you know the distance plays a role in the equation for calculating the gravitational pull $F = G cdot m_1 cdot m_2 cdot 1/r^2$.



However, the difference is normally so small that it cannot be noticed without sensitive instruments. When it becomes noticeable it means that tidal forces are so strong that the body is about to be disrupted.






share|improve this answer











$endgroup$












  • $begingroup$
    This answer is begging to mention tides, which (at measurable levels) makes the measured "gravity" towards the center of the planet equal at b and a, and less than the gravity at the top and bottom of the planet.
    $endgroup$
    – Yakk
    4 mins ago










  • $begingroup$
    For the sake of inclusion, here's the equation for finding the effect of Gravity. (if you don't want to click the link it's: G * m1 * m2 / r^2)
    $endgroup$
    – David
    3 mins ago











  • $begingroup$
    @Yakk, an entire ocean to notice a shift of few meters (at most) falls into my definition of "sensitive instrument"
    $endgroup$
    – L.Dutch
    18 secs ago


















2












$begingroup$

Not easily.



An orbiting object has the feature of being in free fall, which means that objects on it are not subject to gravitational pull from the object that is being orbited, no more than a person inside an orbiting space station feel no gravitational pull from the Earth.



There is however such a thing as tidal forces. Since all parts of a planet orbits the sun at the same orbital speed, even though only the centre of mass is fulfilling orbital equations, objects on the near side experience more pull from the sun that objects on the far side, since the near side moves slower than an object in circular orbit at that distance from the sun would, while the far side moves faster than an object at that side would. This is typically too little to be noticeable by people on Earth, though one-third of the tidal forecs creating ocean tides stems from the sun.



To make it word as you want, you would need the planet to be either far closer to its sun, compared to its mass, or being far bigger (creating a greater tidal difference between near and far side). If the planet has slow rotation, e.g. only having one day and night per orbit, it may be experienced as having lower gravity during the day than during the night. Or if the planet has a very eccentric orbit, the gravity differences between night and day will be experienced as greater the closer the planet is to the sun.



The major problem here is that any tidal force strong enough to be felt as noticably different gravitational pull will very likely tear the planet apart, not to mention creating extremely high and low tides. The latter may not be a problem if the entire surface of the planet is ocean (or if the planet is very plastic), since the inhabitants would move with the surface. Another result of such high gravitational pull is however that the planet will most likely be tidally locked, meaning it will always have the same tide facing the sun, same as our moon always has the same side facing the Earth.



You would need a planet that is made of extremely strong and/or very plastic material to make your idea work, and even then, there are likely to be associated problems such as extreme tides and extreme temperature differences between day and night.






share|improve this answer









$endgroup$




















    1












    $begingroup$

    I can think of a somewhat contrived way that this can happen, but it will be a smaller effect than you'd probably like, and with severe restrictions on the planet's climate and types of civilizations that could live there.



    The idea is that your planet could be in a highly elliptical orbit that causes it to experience large temperature changes throughout the year. This in turn could cause the planet to expand and contract, thus changing its surface gravity (i.e. when the planet is closer to the sun, it would have a lower gravity on the surface, because its surface is physically farther away from its own center of mass).



    Thermal expansion alone won't get you this effect on a rocky planet - solid volumes tend to increase only by about 1% at most even when subjected to temperature changes of 100 K (which is where you should probably draw the line as far as this planet being habitable is concerned). One workaround could be for the planet to have pockets of gas throughout its interior; when the planet heats up, pressure from those pockets could inflate the entire planet like a balloon. However, I'm doubtful that you could get more than a percent or two from this either, and it also comes with more details that need to be addressed, like why geological activity doesn't push most of this gas to the surface.



    No, if you want to go the thermal expansion route, the best way I can think of is to have floating cities above a low-mass, gaseous planet (a gas dwarf). This is all conjecture since there are no confirmed gas dwarf exoplanets, but one with the mass of Earth, similar to Kepler-138d, could ostensibly be 80% gas by volume (60% by radius). If this planet's average temperature were to range from 250 K to 300 K according to the time of year, let's say, then its radius would increase by about 4% and its gravity would decrease by roughly 7%.*



    *These are back-of-the-envelope calculations assuming isobaric expansion, which isn't entirely valid when you're dealing with planet-scale gravitational effects. But the volume difference is only 20%, so it's probably close enough.



    Not a huge effect, but definitely noticeable. Could be an interesting way to have the effect you're looking for, though the sky-city-on-a-gas-planet thing might be too specific for the setting you have in mind.






    share|improve this answer








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    $endgroup$




















      0












      $begingroup$

      You could change the distance relation of the gravitational force field. In the real world it decreases quadratic with distance F=F(r^2) but you can change that to any function you like. Something higher order like r^3 and above will have more turning points. You could put your planet in a distant of such a turning point and depending on your coefficients the resulting gravitation on the closer to further sunside would be different. Does that make sense?





      share








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        Your Answer





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        4 Answers
        4






        active

        oldest

        votes








        4 Answers
        4






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        5












        $begingroup$

        It already happens.



        Look at the schematic drawing below, where you see a planet and its Sun (not to scale), with two different places on the planet: one facing the Sun, the other on the opposite side from it.



        schematic of the gravitational set up



        Gravitational pull can be schematized with a vector, and the resulting pull is the result of the summation of all the pulls in a given spot.



        In the place b in the picture, the pulls of the Sun and of the planet will operate in the same direction, while in a the pull from the Sun will partially counter the pull from the planet.



        Therefore in b the pull will be stronger than in a.



        Also, during the course of an orbit around the Sun, the distance between the two bodies changes slightly, and as you know the distance plays a role in the equation for calculating the gravitational pull $F = G cdot m_1 cdot m_2 cdot 1/r^2$.



        However, the difference is normally so small that it cannot be noticed without sensitive instruments. When it becomes noticeable it means that tidal forces are so strong that the body is about to be disrupted.






        share|improve this answer











        $endgroup$












        • $begingroup$
          This answer is begging to mention tides, which (at measurable levels) makes the measured "gravity" towards the center of the planet equal at b and a, and less than the gravity at the top and bottom of the planet.
          $endgroup$
          – Yakk
          4 mins ago










        • $begingroup$
          For the sake of inclusion, here's the equation for finding the effect of Gravity. (if you don't want to click the link it's: G * m1 * m2 / r^2)
          $endgroup$
          – David
          3 mins ago











        • $begingroup$
          @Yakk, an entire ocean to notice a shift of few meters (at most) falls into my definition of "sensitive instrument"
          $endgroup$
          – L.Dutch
          18 secs ago















        5












        $begingroup$

        It already happens.



        Look at the schematic drawing below, where you see a planet and its Sun (not to scale), with two different places on the planet: one facing the Sun, the other on the opposite side from it.



        schematic of the gravitational set up



        Gravitational pull can be schematized with a vector, and the resulting pull is the result of the summation of all the pulls in a given spot.



        In the place b in the picture, the pulls of the Sun and of the planet will operate in the same direction, while in a the pull from the Sun will partially counter the pull from the planet.



        Therefore in b the pull will be stronger than in a.



        Also, during the course of an orbit around the Sun, the distance between the two bodies changes slightly, and as you know the distance plays a role in the equation for calculating the gravitational pull $F = G cdot m_1 cdot m_2 cdot 1/r^2$.



        However, the difference is normally so small that it cannot be noticed without sensitive instruments. When it becomes noticeable it means that tidal forces are so strong that the body is about to be disrupted.






        share|improve this answer











        $endgroup$












        • $begingroup$
          This answer is begging to mention tides, which (at measurable levels) makes the measured "gravity" towards the center of the planet equal at b and a, and less than the gravity at the top and bottom of the planet.
          $endgroup$
          – Yakk
          4 mins ago










        • $begingroup$
          For the sake of inclusion, here's the equation for finding the effect of Gravity. (if you don't want to click the link it's: G * m1 * m2 / r^2)
          $endgroup$
          – David
          3 mins ago











        • $begingroup$
          @Yakk, an entire ocean to notice a shift of few meters (at most) falls into my definition of "sensitive instrument"
          $endgroup$
          – L.Dutch
          18 secs ago













        5












        5








        5





        $begingroup$

        It already happens.



        Look at the schematic drawing below, where you see a planet and its Sun (not to scale), with two different places on the planet: one facing the Sun, the other on the opposite side from it.



        schematic of the gravitational set up



        Gravitational pull can be schematized with a vector, and the resulting pull is the result of the summation of all the pulls in a given spot.



        In the place b in the picture, the pulls of the Sun and of the planet will operate in the same direction, while in a the pull from the Sun will partially counter the pull from the planet.



        Therefore in b the pull will be stronger than in a.



        Also, during the course of an orbit around the Sun, the distance between the two bodies changes slightly, and as you know the distance plays a role in the equation for calculating the gravitational pull $F = G cdot m_1 cdot m_2 cdot 1/r^2$.



        However, the difference is normally so small that it cannot be noticed without sensitive instruments. When it becomes noticeable it means that tidal forces are so strong that the body is about to be disrupted.






        share|improve this answer











        $endgroup$



        It already happens.



        Look at the schematic drawing below, where you see a planet and its Sun (not to scale), with two different places on the planet: one facing the Sun, the other on the opposite side from it.



        schematic of the gravitational set up



        Gravitational pull can be schematized with a vector, and the resulting pull is the result of the summation of all the pulls in a given spot.



        In the place b in the picture, the pulls of the Sun and of the planet will operate in the same direction, while in a the pull from the Sun will partially counter the pull from the planet.



        Therefore in b the pull will be stronger than in a.



        Also, during the course of an orbit around the Sun, the distance between the two bodies changes slightly, and as you know the distance plays a role in the equation for calculating the gravitational pull $F = G cdot m_1 cdot m_2 cdot 1/r^2$.



        However, the difference is normally so small that it cannot be noticed without sensitive instruments. When it becomes noticeable it means that tidal forces are so strong that the body is about to be disrupted.







        share|improve this answer














        share|improve this answer



        share|improve this answer








        edited 1 min ago

























        answered 5 hours ago









        L.DutchL.Dutch

        90.6k29210437




        90.6k29210437











        • $begingroup$
          This answer is begging to mention tides, which (at measurable levels) makes the measured "gravity" towards the center of the planet equal at b and a, and less than the gravity at the top and bottom of the planet.
          $endgroup$
          – Yakk
          4 mins ago










        • $begingroup$
          For the sake of inclusion, here's the equation for finding the effect of Gravity. (if you don't want to click the link it's: G * m1 * m2 / r^2)
          $endgroup$
          – David
          3 mins ago











        • $begingroup$
          @Yakk, an entire ocean to notice a shift of few meters (at most) falls into my definition of "sensitive instrument"
          $endgroup$
          – L.Dutch
          18 secs ago
















        • $begingroup$
          This answer is begging to mention tides, which (at measurable levels) makes the measured "gravity" towards the center of the planet equal at b and a, and less than the gravity at the top and bottom of the planet.
          $endgroup$
          – Yakk
          4 mins ago










        • $begingroup$
          For the sake of inclusion, here's the equation for finding the effect of Gravity. (if you don't want to click the link it's: G * m1 * m2 / r^2)
          $endgroup$
          – David
          3 mins ago











        • $begingroup$
          @Yakk, an entire ocean to notice a shift of few meters (at most) falls into my definition of "sensitive instrument"
          $endgroup$
          – L.Dutch
          18 secs ago















        $begingroup$
        This answer is begging to mention tides, which (at measurable levels) makes the measured "gravity" towards the center of the planet equal at b and a, and less than the gravity at the top and bottom of the planet.
        $endgroup$
        – Yakk
        4 mins ago




        $begingroup$
        This answer is begging to mention tides, which (at measurable levels) makes the measured "gravity" towards the center of the planet equal at b and a, and less than the gravity at the top and bottom of the planet.
        $endgroup$
        – Yakk
        4 mins ago












        $begingroup$
        For the sake of inclusion, here's the equation for finding the effect of Gravity. (if you don't want to click the link it's: G * m1 * m2 / r^2)
        $endgroup$
        – David
        3 mins ago





        $begingroup$
        For the sake of inclusion, here's the equation for finding the effect of Gravity. (if you don't want to click the link it's: G * m1 * m2 / r^2)
        $endgroup$
        – David
        3 mins ago













        $begingroup$
        @Yakk, an entire ocean to notice a shift of few meters (at most) falls into my definition of "sensitive instrument"
        $endgroup$
        – L.Dutch
        18 secs ago




        $begingroup$
        @Yakk, an entire ocean to notice a shift of few meters (at most) falls into my definition of "sensitive instrument"
        $endgroup$
        – L.Dutch
        18 secs ago











        2












        $begingroup$

        Not easily.



        An orbiting object has the feature of being in free fall, which means that objects on it are not subject to gravitational pull from the object that is being orbited, no more than a person inside an orbiting space station feel no gravitational pull from the Earth.



        There is however such a thing as tidal forces. Since all parts of a planet orbits the sun at the same orbital speed, even though only the centre of mass is fulfilling orbital equations, objects on the near side experience more pull from the sun that objects on the far side, since the near side moves slower than an object in circular orbit at that distance from the sun would, while the far side moves faster than an object at that side would. This is typically too little to be noticeable by people on Earth, though one-third of the tidal forecs creating ocean tides stems from the sun.



        To make it word as you want, you would need the planet to be either far closer to its sun, compared to its mass, or being far bigger (creating a greater tidal difference between near and far side). If the planet has slow rotation, e.g. only having one day and night per orbit, it may be experienced as having lower gravity during the day than during the night. Or if the planet has a very eccentric orbit, the gravity differences between night and day will be experienced as greater the closer the planet is to the sun.



        The major problem here is that any tidal force strong enough to be felt as noticably different gravitational pull will very likely tear the planet apart, not to mention creating extremely high and low tides. The latter may not be a problem if the entire surface of the planet is ocean (or if the planet is very plastic), since the inhabitants would move with the surface. Another result of such high gravitational pull is however that the planet will most likely be tidally locked, meaning it will always have the same tide facing the sun, same as our moon always has the same side facing the Earth.



        You would need a planet that is made of extremely strong and/or very plastic material to make your idea work, and even then, there are likely to be associated problems such as extreme tides and extreme temperature differences between day and night.






        share|improve this answer









        $endgroup$

















          2












          $begingroup$

          Not easily.



          An orbiting object has the feature of being in free fall, which means that objects on it are not subject to gravitational pull from the object that is being orbited, no more than a person inside an orbiting space station feel no gravitational pull from the Earth.



          There is however such a thing as tidal forces. Since all parts of a planet orbits the sun at the same orbital speed, even though only the centre of mass is fulfilling orbital equations, objects on the near side experience more pull from the sun that objects on the far side, since the near side moves slower than an object in circular orbit at that distance from the sun would, while the far side moves faster than an object at that side would. This is typically too little to be noticeable by people on Earth, though one-third of the tidal forecs creating ocean tides stems from the sun.



          To make it word as you want, you would need the planet to be either far closer to its sun, compared to its mass, or being far bigger (creating a greater tidal difference between near and far side). If the planet has slow rotation, e.g. only having one day and night per orbit, it may be experienced as having lower gravity during the day than during the night. Or if the planet has a very eccentric orbit, the gravity differences between night and day will be experienced as greater the closer the planet is to the sun.



          The major problem here is that any tidal force strong enough to be felt as noticably different gravitational pull will very likely tear the planet apart, not to mention creating extremely high and low tides. The latter may not be a problem if the entire surface of the planet is ocean (or if the planet is very plastic), since the inhabitants would move with the surface. Another result of such high gravitational pull is however that the planet will most likely be tidally locked, meaning it will always have the same tide facing the sun, same as our moon always has the same side facing the Earth.



          You would need a planet that is made of extremely strong and/or very plastic material to make your idea work, and even then, there are likely to be associated problems such as extreme tides and extreme temperature differences between day and night.






          share|improve this answer









          $endgroup$















            2












            2








            2





            $begingroup$

            Not easily.



            An orbiting object has the feature of being in free fall, which means that objects on it are not subject to gravitational pull from the object that is being orbited, no more than a person inside an orbiting space station feel no gravitational pull from the Earth.



            There is however such a thing as tidal forces. Since all parts of a planet orbits the sun at the same orbital speed, even though only the centre of mass is fulfilling orbital equations, objects on the near side experience more pull from the sun that objects on the far side, since the near side moves slower than an object in circular orbit at that distance from the sun would, while the far side moves faster than an object at that side would. This is typically too little to be noticeable by people on Earth, though one-third of the tidal forecs creating ocean tides stems from the sun.



            To make it word as you want, you would need the planet to be either far closer to its sun, compared to its mass, or being far bigger (creating a greater tidal difference between near and far side). If the planet has slow rotation, e.g. only having one day and night per orbit, it may be experienced as having lower gravity during the day than during the night. Or if the planet has a very eccentric orbit, the gravity differences between night and day will be experienced as greater the closer the planet is to the sun.



            The major problem here is that any tidal force strong enough to be felt as noticably different gravitational pull will very likely tear the planet apart, not to mention creating extremely high and low tides. The latter may not be a problem if the entire surface of the planet is ocean (or if the planet is very plastic), since the inhabitants would move with the surface. Another result of such high gravitational pull is however that the planet will most likely be tidally locked, meaning it will always have the same tide facing the sun, same as our moon always has the same side facing the Earth.



            You would need a planet that is made of extremely strong and/or very plastic material to make your idea work, and even then, there are likely to be associated problems such as extreme tides and extreme temperature differences between day and night.






            share|improve this answer









            $endgroup$



            Not easily.



            An orbiting object has the feature of being in free fall, which means that objects on it are not subject to gravitational pull from the object that is being orbited, no more than a person inside an orbiting space station feel no gravitational pull from the Earth.



            There is however such a thing as tidal forces. Since all parts of a planet orbits the sun at the same orbital speed, even though only the centre of mass is fulfilling orbital equations, objects on the near side experience more pull from the sun that objects on the far side, since the near side moves slower than an object in circular orbit at that distance from the sun would, while the far side moves faster than an object at that side would. This is typically too little to be noticeable by people on Earth, though one-third of the tidal forecs creating ocean tides stems from the sun.



            To make it word as you want, you would need the planet to be either far closer to its sun, compared to its mass, or being far bigger (creating a greater tidal difference between near and far side). If the planet has slow rotation, e.g. only having one day and night per orbit, it may be experienced as having lower gravity during the day than during the night. Or if the planet has a very eccentric orbit, the gravity differences between night and day will be experienced as greater the closer the planet is to the sun.



            The major problem here is that any tidal force strong enough to be felt as noticably different gravitational pull will very likely tear the planet apart, not to mention creating extremely high and low tides. The latter may not be a problem if the entire surface of the planet is ocean (or if the planet is very plastic), since the inhabitants would move with the surface. Another result of such high gravitational pull is however that the planet will most likely be tidally locked, meaning it will always have the same tide facing the sun, same as our moon always has the same side facing the Earth.



            You would need a planet that is made of extremely strong and/or very plastic material to make your idea work, and even then, there are likely to be associated problems such as extreme tides and extreme temperature differences between day and night.







            share|improve this answer












            share|improve this answer



            share|improve this answer










            answered 5 hours ago









            Klaus Æ. MogensenKlaus Æ. Mogensen

            879135




            879135





















                1












                $begingroup$

                I can think of a somewhat contrived way that this can happen, but it will be a smaller effect than you'd probably like, and with severe restrictions on the planet's climate and types of civilizations that could live there.



                The idea is that your planet could be in a highly elliptical orbit that causes it to experience large temperature changes throughout the year. This in turn could cause the planet to expand and contract, thus changing its surface gravity (i.e. when the planet is closer to the sun, it would have a lower gravity on the surface, because its surface is physically farther away from its own center of mass).



                Thermal expansion alone won't get you this effect on a rocky planet - solid volumes tend to increase only by about 1% at most even when subjected to temperature changes of 100 K (which is where you should probably draw the line as far as this planet being habitable is concerned). One workaround could be for the planet to have pockets of gas throughout its interior; when the planet heats up, pressure from those pockets could inflate the entire planet like a balloon. However, I'm doubtful that you could get more than a percent or two from this either, and it also comes with more details that need to be addressed, like why geological activity doesn't push most of this gas to the surface.



                No, if you want to go the thermal expansion route, the best way I can think of is to have floating cities above a low-mass, gaseous planet (a gas dwarf). This is all conjecture since there are no confirmed gas dwarf exoplanets, but one with the mass of Earth, similar to Kepler-138d, could ostensibly be 80% gas by volume (60% by radius). If this planet's average temperature were to range from 250 K to 300 K according to the time of year, let's say, then its radius would increase by about 4% and its gravity would decrease by roughly 7%.*



                *These are back-of-the-envelope calculations assuming isobaric expansion, which isn't entirely valid when you're dealing with planet-scale gravitational effects. But the volume difference is only 20%, so it's probably close enough.



                Not a huge effect, but definitely noticeable. Could be an interesting way to have the effect you're looking for, though the sky-city-on-a-gas-planet thing might be too specific for the setting you have in mind.






                share|improve this answer








                New contributor




                Gilad M is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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                $endgroup$

















                  1












                  $begingroup$

                  I can think of a somewhat contrived way that this can happen, but it will be a smaller effect than you'd probably like, and with severe restrictions on the planet's climate and types of civilizations that could live there.



                  The idea is that your planet could be in a highly elliptical orbit that causes it to experience large temperature changes throughout the year. This in turn could cause the planet to expand and contract, thus changing its surface gravity (i.e. when the planet is closer to the sun, it would have a lower gravity on the surface, because its surface is physically farther away from its own center of mass).



                  Thermal expansion alone won't get you this effect on a rocky planet - solid volumes tend to increase only by about 1% at most even when subjected to temperature changes of 100 K (which is where you should probably draw the line as far as this planet being habitable is concerned). One workaround could be for the planet to have pockets of gas throughout its interior; when the planet heats up, pressure from those pockets could inflate the entire planet like a balloon. However, I'm doubtful that you could get more than a percent or two from this either, and it also comes with more details that need to be addressed, like why geological activity doesn't push most of this gas to the surface.



                  No, if you want to go the thermal expansion route, the best way I can think of is to have floating cities above a low-mass, gaseous planet (a gas dwarf). This is all conjecture since there are no confirmed gas dwarf exoplanets, but one with the mass of Earth, similar to Kepler-138d, could ostensibly be 80% gas by volume (60% by radius). If this planet's average temperature were to range from 250 K to 300 K according to the time of year, let's say, then its radius would increase by about 4% and its gravity would decrease by roughly 7%.*



                  *These are back-of-the-envelope calculations assuming isobaric expansion, which isn't entirely valid when you're dealing with planet-scale gravitational effects. But the volume difference is only 20%, so it's probably close enough.



                  Not a huge effect, but definitely noticeable. Could be an interesting way to have the effect you're looking for, though the sky-city-on-a-gas-planet thing might be too specific for the setting you have in mind.






                  share|improve this answer








                  New contributor




                  Gilad M is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                  Check out our Code of Conduct.






                  $endgroup$















                    1












                    1








                    1





                    $begingroup$

                    I can think of a somewhat contrived way that this can happen, but it will be a smaller effect than you'd probably like, and with severe restrictions on the planet's climate and types of civilizations that could live there.



                    The idea is that your planet could be in a highly elliptical orbit that causes it to experience large temperature changes throughout the year. This in turn could cause the planet to expand and contract, thus changing its surface gravity (i.e. when the planet is closer to the sun, it would have a lower gravity on the surface, because its surface is physically farther away from its own center of mass).



                    Thermal expansion alone won't get you this effect on a rocky planet - solid volumes tend to increase only by about 1% at most even when subjected to temperature changes of 100 K (which is where you should probably draw the line as far as this planet being habitable is concerned). One workaround could be for the planet to have pockets of gas throughout its interior; when the planet heats up, pressure from those pockets could inflate the entire planet like a balloon. However, I'm doubtful that you could get more than a percent or two from this either, and it also comes with more details that need to be addressed, like why geological activity doesn't push most of this gas to the surface.



                    No, if you want to go the thermal expansion route, the best way I can think of is to have floating cities above a low-mass, gaseous planet (a gas dwarf). This is all conjecture since there are no confirmed gas dwarf exoplanets, but one with the mass of Earth, similar to Kepler-138d, could ostensibly be 80% gas by volume (60% by radius). If this planet's average temperature were to range from 250 K to 300 K according to the time of year, let's say, then its radius would increase by about 4% and its gravity would decrease by roughly 7%.*



                    *These are back-of-the-envelope calculations assuming isobaric expansion, which isn't entirely valid when you're dealing with planet-scale gravitational effects. But the volume difference is only 20%, so it's probably close enough.



                    Not a huge effect, but definitely noticeable. Could be an interesting way to have the effect you're looking for, though the sky-city-on-a-gas-planet thing might be too specific for the setting you have in mind.






                    share|improve this answer








                    New contributor




                    Gilad M is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                    Check out our Code of Conduct.






                    $endgroup$



                    I can think of a somewhat contrived way that this can happen, but it will be a smaller effect than you'd probably like, and with severe restrictions on the planet's climate and types of civilizations that could live there.



                    The idea is that your planet could be in a highly elliptical orbit that causes it to experience large temperature changes throughout the year. This in turn could cause the planet to expand and contract, thus changing its surface gravity (i.e. when the planet is closer to the sun, it would have a lower gravity on the surface, because its surface is physically farther away from its own center of mass).



                    Thermal expansion alone won't get you this effect on a rocky planet - solid volumes tend to increase only by about 1% at most even when subjected to temperature changes of 100 K (which is where you should probably draw the line as far as this planet being habitable is concerned). One workaround could be for the planet to have pockets of gas throughout its interior; when the planet heats up, pressure from those pockets could inflate the entire planet like a balloon. However, I'm doubtful that you could get more than a percent or two from this either, and it also comes with more details that need to be addressed, like why geological activity doesn't push most of this gas to the surface.



                    No, if you want to go the thermal expansion route, the best way I can think of is to have floating cities above a low-mass, gaseous planet (a gas dwarf). This is all conjecture since there are no confirmed gas dwarf exoplanets, but one with the mass of Earth, similar to Kepler-138d, could ostensibly be 80% gas by volume (60% by radius). If this planet's average temperature were to range from 250 K to 300 K according to the time of year, let's say, then its radius would increase by about 4% and its gravity would decrease by roughly 7%.*



                    *These are back-of-the-envelope calculations assuming isobaric expansion, which isn't entirely valid when you're dealing with planet-scale gravitational effects. But the volume difference is only 20%, so it's probably close enough.



                    Not a huge effect, but definitely noticeable. Could be an interesting way to have the effect you're looking for, though the sky-city-on-a-gas-planet thing might be too specific for the setting you have in mind.







                    share|improve this answer








                    New contributor




                    Gilad M is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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                    share|improve this answer



                    share|improve this answer






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                    answered 3 hours ago









                    Gilad MGilad M

                    413




                    413




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                        0












                        $begingroup$

                        You could change the distance relation of the gravitational force field. In the real world it decreases quadratic with distance F=F(r^2) but you can change that to any function you like. Something higher order like r^3 and above will have more turning points. You could put your planet in a distant of such a turning point and depending on your coefficients the resulting gravitation on the closer to further sunside would be different. Does that make sense?





                        share








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                          0












                          $begingroup$

                          You could change the distance relation of the gravitational force field. In the real world it decreases quadratic with distance F=F(r^2) but you can change that to any function you like. Something higher order like r^3 and above will have more turning points. You could put your planet in a distant of such a turning point and depending on your coefficients the resulting gravitation on the closer to further sunside would be different. Does that make sense?





                          share








                          New contributor




                          Martin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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                          $endgroup$















                            0












                            0








                            0





                            $begingroup$

                            You could change the distance relation of the gravitational force field. In the real world it decreases quadratic with distance F=F(r^2) but you can change that to any function you like. Something higher order like r^3 and above will have more turning points. You could put your planet in a distant of such a turning point and depending on your coefficients the resulting gravitation on the closer to further sunside would be different. Does that make sense?





                            share








                            New contributor




                            Martin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                            Check out our Code of Conduct.






                            $endgroup$



                            You could change the distance relation of the gravitational force field. In the real world it decreases quadratic with distance F=F(r^2) but you can change that to any function you like. Something higher order like r^3 and above will have more turning points. You could put your planet in a distant of such a turning point and depending on your coefficients the resulting gravitation on the closer to further sunside would be different. Does that make sense?






                            share








                            New contributor




                            Martin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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                            share


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                            answered 9 mins ago









                            MartinMartin

                            1




                            1




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