What is the magnitude of the free-fall acceleration at a point that is a distance 2R above the surface of the Earth, where R is

Question

3 years ago

9

the radius of the Earth

1 answer:

ss7ja [257] · Answer 1 · 2022-06-04T07:57:10+03:00

Answer:

g' = g/9 = 1.09 m/s²

Explanation:

The magnitude of free fall acceleration at the surface of earth is given by the following formula:

g = GM/R² ----- equation 1

where,

g = free fall acceleration

G = Universal Gravitational Constant

M = Mass of Earth

R = Distance between the center of earth and the object

So, in our case,

R = R + 2 R = 3 R

Therefore,

g' = GM/(3R)²

g' = (1/9) GM/R²

using equation 1:

g' = g/9

g' = (9.8 m/s)/9