Question

If a planet has the same mass as the earth, but has twice the radius, how does the surface gravity, g, compare to g on the surface of the earth

Answer 1

The surface gravity g of the planet is 1/4 of the surface gravity on earth.

Step-by-step explanation:

Surface gravity is given by the following formula:

`g=G(m)/(r^(2))`

So the gravity of both the earth and the planet is written in terms of their own radius, so we get:

`g_(E)=G(m)/(r_(E)^(2))`

`g_(P)=G(m)/(r_(P)^(2))`

The problem tells us the radius of the planet is twice that of the radius on earth, so:

`r_(P)=2r_(E)`

If we substituted that into the gravity of the planet equation we would end up with the following formula:

`g_(P)=G(m)/((2r_(E))^(2))`

Which yields:

`g_(P)=G(m)/(4r_(E)^(2))`

So we can now compare the two gravities:

`(g_(P))/(g_(E))=(G(m)/(4r_(E)^(2)))/(G(m)/(r_(E)^(2)))`

When simplifying the ratio we end up with:

`(g_(P))/(g_(E))=(1)/(4)`

So the gravity acceleration on the surface of the planet is 1/4 of that on the surface of Earth.