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?

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

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)/(()2

x = 7×4 GM/r2    

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

x = 7×4×10 = 280 m/s2

r = ρ x length / area  

ρ = r x area / length  

ρ is in ohm-meters  

in basic units, it is m³kg/s³a²  

i think this is it.

amplitude

hope this