Question

What is the force of gravity (from the Earth) on the 700kg satellite if it’s 10km above the Earth's surface?

Answer 1

Answer:

g = 4.7 × $10^{-16}$ m/$s^{2}$

Explanation:

Given that the mass of the satellite = 700 kg, and 10,000 m above the earth;s surface.

From Newton's second law,

 F = mg ............... 1

From Newton's gravitation law,

F = $\frac{GMm}{r^{2} }$ .................. 2

Where: F is the force, G is the gravitational constant, M is the mass of the first body, m is the mass of the second body, g is the gravitational force and r is the distance between the centers of the two bodies.

Equate 1 and 2 to have,

mg = $\frac{GMm}{r^{2} }$

⇒   g = $\frac{GM}{r^{2} }$

But; G =  6.67 × $10^{-11}$ N $m^{2} Kg^{-2}$, M = 700 Kg, r = 10000 m

Thus,

g = $\frac{6.67*10^{-11*700} }{10000^{2} }$

  = $\frac{4.669*10^{-8} }{1*10^{8} }$

  = 4.669 × $10^{-16}$ m/$s^{2}$

The force of gravity on the satellite is 4.7 × $10^{-16}$ m/$s^{2}$.