A Salute to Isaac Newton

Classical Mechanics Level 1

If the Earth's mass and radius are both doubled, then what will happen to the gravitational force felt by an object on its surface?

It will double. It won't change. It will be halved. It will quadruple.

1 solution

Syed Baqir
Sep 4, 2015

Suppose that the Earth's mass and radius are given by $M_e$ and $r_e$ respectively. Then an object (of mass $m$ ) on its surface will feel a gravitational force given by $F = \frac{GM_em}{r_{e}^{2}}.$

If we double the mass and the radius of the Earth, then the gravitational force felt by an object on the surface is now given by $F = \frac{G(2M_e)m}{(2r_e)^2} = \frac{2GM_em}{4r_{e}^{2}} = \color{#D61F06}{\frac{1}{2}}\frac{GM_em}{r_{e}^{2}}.$

This force is half of the original force.

I cannot understand, aren't a planet bigger in size supposed to have larger gravity...it must make a larger curvature in the continuum.

Ahmed Taha - 5 years, 9 months ago

They are, but two times bigger planets aren't expected to have double the mass, but more than that. Actually, if the two times bigger planet was made of the same material as a smaller one (which isn't always the case, there are gaseous giants vs. smaller rocky planets), the volume would be eight times as large, thus giving 8 times more mass. HINT: Volume of a sphere = ...r^3

Alija Bevrnja - 5 years, 9 months ago

I understood

Sonal Singh - 5 years, 9 months ago

Thanks by the answer! =D

Cris Contin - 5 years, 9 months ago

Obviously your answer is correct, but I'm just curious. What are you using for the second mass in the gravitational equation between two objects? The equation g=(GM)/r^2 could be used since it only has only one variable for mass, but it calculates acceleration instead of force.

Blaine Gustafson - 5 years, 9 months ago

The question asked as about gravitational Force , not gravitational acceleration !

you could use GM/r^2 if you are asked for acceleration.

Syed Baqir - 5 years, 9 months ago

(and Blaine Gustafson): Well it doesn't make difference, because only the ratio of new vs. old force is sought. Hence, whatever the second mass, it will divide out in the end. The ratio is what matters and it will be the same for the accelerations.

Alija Bevrnja - 5 years, 9 months ago

A Salute to Isaac Newton

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