For example, a grasshopper is perched on a blade of grass on the Earth. Logically, I would say the earth exerts greater gravitational force due to its mass but is this correct?

If not, which exerts the greater gravitational force, the grasshopper on the earth or the earth on the grasshopper?


Whenever gravity exerts a force on an object, it is because there is another object in the vicinity. Both masses are required to create the force. If the mass of one object is double, the force is doubled. This happens for either object.

The formula for the gravitational force between the Earth (E) and the grasshopper (g) is $$F_{gE} = \frac{Gm_gm_E}{r^2}$$ where $F_{gE}$ is the force on the grasshopper due to the Earth, $m_g$ is the mass of the grasshopper, $m_E$ is the mass of the Earth, $r$S is the distance between the grasshopper and the center of the Earth, and $G$ is a constant.

Now, lets find the force on the Earth due to the grasshopper. We just switch the roles of $E$ and $g$: $$F_{Eg} = \frac{Gm_Em_g}{r^2}.$$ Because, $m_gm_E = m_Em_g$, we see that the force is the same. The grasshopper pulls on the Earth just as hard as the Earth pulls on the grasshopper.

This is one example of Newton's third law in action. If A exerts a force on B, the B exerts a force of equal magnitude on A.


The gravitational force between two bodies is


Where F is the force, m1 and m2 are the masses of the objects interacting, r is the distance between the centers of the masses and G is the gravitational constant.

So in principle larger bodies give a larger contribution to the force between m1 and m2, but the size of the force that each feels is the same , by Newton's third law.

The F=mg used to calculate the trajectories of bodies at the surface of the earth is a valid approximation due to the very large numbers involved , some numbers here.

  • 2
    $\begingroup$ It looks to me like the product of m1 and m2 does not distinguish between the masses. $\endgroup$ – C. Towne Springer Feb 4 '17 at 5:55
  • $\begingroup$ @C.TowneSpringer m1 and m2 have a size, a number in the field of real numbers. one may be 10^16 and the other 1. My mistaken formulation. OK I will edit to "give rise to". $\endgroup$ – anna v Feb 4 '17 at 6:57
  • $\begingroup$ I couldn't see anything what is wrong in this answer. It would be a good practice to give a comment for downvotes. $\endgroup$ – HolgerFiedler Feb 4 '17 at 11:25
  • $\begingroup$ @HolgerFiedler the way It was written before the edit I was breakin Newton's third law ( too soon after waking up :) $\endgroup$ – anna v Feb 4 '17 at 12:10

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