I was curious about what my weight would be in different planets so I found myself on this site:


But while Neptune is about 17 times more massive than earth, if you were to place some scales on Neptune and stand on them, you would not be 17 times your earth weight. Why is this the case?


The weight of an object $W$ of mass $m$ at the surface of a planet of radius $R$ and mass $M$ can be found by using Newton's law of gravitation.

$W = \dfrac {GMm}{R^2}$ where $G$ is the gravitational constant.

So $\dfrac{W_{\rm Neptune}}{W_{\rm Earth}}=\left( \dfrac{M_{\rm Neptune}}{M_{\rm Earth}}\right)\left( \dfrac{R_{\rm Earth}}{R_{\rm Neptune}}\right)^2$

Using tabulated data produced by NASA gives

$\dfrac{W_{\rm Neptune}}{W_{\rm Earth}}=17.1\left(\dfrac{1}{3.88} \right)^2 = 1.14 $

Approximations have been made to produce this figure in that the planets were assume spherical, not rotating etc.


EDIT This answer is incorrect (to say the least) due to hasty assumptions on my part and I am very grateful to Albert Aspect for his comment below, and the link provided by him that I should have consulted before attempting an answer. Gauss's Theorem. END EDIT

The radius of the planet is also involved. Neptune is massive, but it's centre is so far away from it's "surface", that the effective pull of gravity on your body is less than you would calculate using the mass of Neptune alone.

The important word in the excerpt below is center.

From Newton's Law of Gravitational Attraction

Every point mass attracts every single other point mass by a force pointing along the line intersecting both points. The force is proportional to the product of the two masses and inversely proportional to the square of the distance between them:

$${\displaystyle F=G{\frac {m_{1}m_{2}}{r^{2}}}\ } $$ where:

$F$ is the force between the masses;

$G$ is the gravitational constant

$m_1$ is the first mass;

$m_2$ is the second mass;

$r$ is the distance between the centers of the masses.

  • $\begingroup$ Does not Gauss theorem contradict your first paragraph? The density distribution as a function of r does not matter if it is spherically symmetrical, only the total mass. hyperphysics.phy-astr.gsu.edu/hbase/Mechanics/sphshell.html $\endgroup$ – user126422 Feb 5 '17 at 5:53
  • $\begingroup$ @AlbertAspect Thank you very much for pointing that out to me (and the politeness/ diplomacy reflected in your comment :) I have edited the answer to reference your comment and include the link you provided, in case your comment is deleted and someone else makes the same incorrect assumption I did. $\endgroup$ – user140606 Feb 5 '17 at 8:16

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