Gravity and Rotation 10?

Question

What is the distance from the Earth's center to a point outside the Earth where the gravitational acceleration due to the Earth is 1/24 of its value at the Earth's surface?

Answer should be in meters.

Answer 1

g = M * G / r^2

g/24 = M * G /r^2

r^2 = M*G * 24 / g

r = sqrt (M*G*24/g)

r = sqrt (5.98 x 10^24 * 24 / 9.81) = 3.82 x 10^12m

So basically what happens is the radius of the earth is multiplied by the sqrt (24).

Answer 2

The force due to gravity is balanced by the gravitational attraction.

mg = GMm/r^2 where g = 1/24 of it's regular value

g/24 = GM/r^2 where r is the distance from the centre of the Earth.

Solve for r, You will need G = 6.67x10^-11 and a value for the mass of the earth

Answer 3

If R is the earth's radius, the distance is R*√24 m