Answer:
The surface gravity g of the planet is 1/4 of the surface gravity on earth.
Explanation:
Surface gravity is given by the following formula:
[tex]g=G\frac{m}{r^{2}}[/tex]
So the gravity of both the earth and the planet is written in terms of their own radius, so we get:
[tex]g_{E}=G\frac{m}{r_{E}^{2}}[/tex]
[tex]g_{P}=G\frac{m}{r_{P}^{2}}[/tex]
The problem tells us the radius of the planet is twice that of the radius on earth, so:
[tex]r_{P}=2r_{E}[/tex]
If we substituted that into the gravity of the planet equation we would end up with the following formula:
[tex]g_{P}=G\frac{m}{(2r_{E})^{2}}[/tex]
Which yields:
[tex]g_{P}=G\frac{m}{4r_{E}^{2}}[/tex]
So we can now compare the two gravities:
[tex]\frac{g_{P}}{g_{E}}=\frac{G\frac{m}{4r_{E}^{2}}}{G\frac{m}{r_{E}^{2}}}[/tex]
When simplifying the ratio we end up with:
[tex]\frac{g_{P}}{g_{E}}=\frac{1}{4}[/tex]
So the gravity acceleration on the surface of the planet is 1/4 of that on the surface of Earth.
