Question

Find the centripetal force of a satellite of mass 1000kg orbiting the earth in a circular orbit close to it's surface. What will be the magnitude of the force if the orbit were not close to the surface of the earth but at a height above it equal to the earth's radius. Why does the centripetal force do not work?

Answer 1

Explanations & Calculations

• The interactive force between two masses is given by Newton's law of gravitation.
• It is the same for a system of earth & an orbiting satellite.
• When the satellite is orbiting the earth closer to the surface of the earth the distance between them can be approximated to the value of the earth's radius $\small 6371\,km$.
• Then for the first part

$\qquad\qquad \begin{aligned} \small F&=\small G\frac{M_em_s}{r^2}\\ &=\small (6.67\times10^{-11}Nkg^{-2}m^2)\cdot\frac{(5.97\times10^{24}kg)\cdot(10^3kg)}{(6371\times10^3m)^2}\\ &=\small \bold{9810.36\,N} \end{aligned}$

• What may happen if the distance between them is doubled

$\qquad\qquad \begin{aligned} \small F_1&=\small G\frac{M_em_s}{(2r)^2}\\ &=\small G\frac{M_em_s}{r^2}\cdot\frac{1}{4}\\ &=\small \frac{F}{4} \end{aligned}$

• The new centripetal force is reduced to a quarter of that when the satellite was near the earth's surface.

• To perform some work by a force, the force's line of action should travel the distance described there. But in a situation like this, the centripetal force\its line of action is perpendicular to the traveling path: circumference.
• Therefore, the performed work is zero

$\qquad\qquad \begin{aligned} \small W&=\small fs\\ &=\small F\cos 90\times2\pi r\\ &=\small 0\times 2\pi r\\ &=\small 0\,J \end{aligned}$