A 60.0 kg mass is located 1500 km above the surface of the Earth. The acceleration due to gravity on the Earth’s surface is g= 9.80 m/s2 and the radius of Earth is 6.37 x 106 m

Question

(a)What is the mass of the 60.0 kg object at an altitude of 1500 km?

(b)What is the acceleration due to gravity at an altitude of 1500 km?

(c)What is the magnitude of the centripetal acceleration of a satellite orbiting at an altitude of 150 0 km?

(d)At what distance above the Earth’s surface would the acceleration due to gravity be 1/5 of its value on the surface of the Earth?

(the answer for a is 60kg, b is 6.42 m/s^2, c is 6.42 m/s^2, d is

7.87 x 10^6 m) im not sure how to get these answers

Answer 1

Let

m = 60 kg be mass of the object,

M (not given) be the mass of the Earth,

R = 6370 km the the radius of the Earth,

h = 1500 km be the distance of the object from the surface,

g = 9.8 m/s² be gravitation acceleration at the surface,

G be the gravitational constant

(a) This is a trick question: Mass (as opposed to weight) is independent of gravity.

(b) In general, gravitational acceleration at a point with distance r from the

center of the Earth is

GM/r².

Therefore

g = GM/R²

This gives us

GM = gR²

This identity will be handy next.

At height h the gravitational acceleration is

GM/(R+h)² = gR²/(R+h)²

In other words, the acceleration at height h is reduced by a factor of R²/(R+h)². (1)

If you plug actual numbers into it, you get the 6.42 m/s².

(c) This is another trick question.

For the satellite to orbit along a circle, its centripetal acceleration

must equal its gravitational acceleration.

The reason is that it is gravity that is causing the centripetal acceleration.

(d) Recall the result (1). We are to calculate x, such that

R²/(R+x)² = 1/5

After simplification you get the quadratic equation

x² + 2Rx - 4R² = 0

The one meaningful solution is

R(sqrt(5)-1)

which gives the provided value.