Suppose that I'm alone on a planet of $M$. From what I learned from school, the gravitation force acting on me is given by


where $m$ is my mass, $r$ is the distance between me and the planet and $G = 6.67 \cdot 10 ^{-11}$.
Suppose that my dog, of mass $\tilde m$, comes to live with me on this planet. Now, is the gravitation force acting on me still $F$ or is it $$\tilde F=G\frac{m(M+\tilde m)}{r^2}~? $$

  • $\begingroup$ What is $r$? Is your dog at the center of the planet? $\endgroup$
    – WillO
    Sep 25, 2015 at 22:14
  • 4
    $\begingroup$ Related: physics.stackexchange.com/q/207854/2451 and links therein. $\endgroup$
    – Qmechanic
    Sep 25, 2015 at 22:16
  • $\begingroup$ @WillO The dog is at the surface. Do the answer depend on the position of the dog? $\endgroup$
    – Surb
    Sep 25, 2015 at 22:16
  • $\begingroup$ @Qmechanic No I have seen this question and it is about more about how to measure the weight of the planet than the principle itself. I would like to understand how the fact that my dog is on the planet change (or not) the attraction $\endgroup$
    – Surb
    Sep 25, 2015 at 22:17
  • $\begingroup$ Why the downvotes? This is a valid question. $\endgroup$
    – CoilKid
    Sep 26, 2015 at 1:09

2 Answers 2


The vertically downward gravitational force exerted by the Earth of mass $M$ on a object of mass $m$ standing on the surface of the Earth is given by:

$F=G\frac{m \times M}{r^2}$ where $r$ is the radius of the Earth.

Now add a dog standing close to you, also on the surface of the Earth, then the dog would exert a (negligible) more or less horizontal force of attraction on you given by:

$F'=G\frac{m \times \tilde m}{d^2}$, where $d$ is the distance between you and the dog.

Since as $F$ and $F'$ are more or less perpendicular to each other (one vertical, one horizontal) the dog does not affect $F$ in the way you describe and:

$\tilde F=G\frac{m(M+\tilde m)}{r^2}$ is not correct. See also the force diagram below:

Force diagram.

One could calculate a resultant force $F_R$ by vector addition of $F$ and $F'$. It would have the magnitude (scalar):

$F_R=\sqrt{F^2+F'^2}$ and with $F \gg F'$, then basically $F_R=F$.


I mean, technically it would affect F, yes. But the mass of your dog is negligible compared to the mass of the planet. Just going off of Earth, we're talking 6e24 kg compared to 50 kg at most. The gravitational force from the planet is so much more than the gravitational force from the dog - which is why in real life we fall towards the Earth, not towards our pets.

EDIT: The effect of the dog would be horizontal, not vertical as Earth's gravitational force is. So OP's (M+m) isn't right, but that doesn't change that we can ignore the dog's mass when calculating gravity.

  • $\begingroup$ So technically $\tilde F$ is correct? $\endgroup$
    – Surb
    Sep 25, 2015 at 22:22
  • 1
    $\begingroup$ For all intents and purposes, it's accurate. Plugging in some numbers for an order of magnitude estimation, a 50 kg dog (a really big dog) that's one meter away from you exerts a gravitational force on you that is 26 orders of magnitude less than the force that Earth exerts on you. $\endgroup$ Sep 25, 2015 at 22:28
  • 1
    $\begingroup$ If you want to get technical, you'd say that the original F was <0,0,10^19> and the new F is <10^-7,0,10^19>, but 10^-7 is small enough to not make a difference in this situation. $\endgroup$ Sep 25, 2015 at 22:29
  • $\begingroup$ Sorry, but this is incorrect. The attractions do not act in the same direction. $\endgroup$
    – Gert
    Sep 25, 2015 at 23:13
  • $\begingroup$ I never said that they did. Look at the vector in the comment. Though I suppose it could be more clear, I'll add an edit to clarify. $\endgroup$ Sep 25, 2015 at 23:23

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