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What would be the acceleration of gravity (in m/s2) at the surface of a world with seven times Earth's mass and half its radius?

User Lixas
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1 Answer

3 votes

Answer:

280 m/s2 if gravitational acceleration upon Earth's surface is 10 m/s2

Step-by-step explanation:

By Newton's law of gravity we know,

F=GMm /r2

and also you know

F = mg

where G = universal gravitational constant

M = Mass of the object 1 (Earth in this case)

m = Mass of the object 2 (any object under the gravitational attraction of Earth )

r = Distance between object 1 and object 2

g = acceleration due to gravity

F = Gravitational force acting on each body

By equating above 2 equations you can get,

mg = F= GMm /r2

mg = GMm /r2

By cancelling m from both sides

g = GM/ r2

Mind G is a constant. Lets take gravitational acceleration upon Earth's surface is 10 m/s2. You get,

10 = GM/ r2 ---------------- (A)

Lets take the gravitational acceleration upon the surface a world with seven times Earth's mass and half its radius as x (or what ever you like)

you get,

x = G(7M)/((
(r)/(2))2

x = 7×4 GM/r2

You know the value of GM/ r2 from (A), by substituting

x = 7×4×10 = 280 m/s2

User Nicolai Harbo
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