Question

A satellite orbits a planet of unknown mass in a circle of radius
2.0×107
m. The magnitude of the gravitational force on the satellite from the planet is F = 80 N.
(a) What is the kinetic energy of the satellite in this orbit?
(b) What would F be if the orbit radius were increased to
3.0×107
m?

Answer 1

Explanation:

It is given that,

Radius of unknown planet, $r=2\times 10^7\ m$

The magnitude of the gravitational force on the satellite from the planet is F = 80 N, it means,

$\dfrac{GmM}{r^2}=80$    

The gravitational force acting on the satellite is balanced by the centripetal force as :

(a) Let E is the is the kinetic energy of the satellite in this orbit.

$\dfrac{GmM}{r^2}=\dfrac{mv^2}{r}$          

$mv^2=\dfrac{GmM}{r^2}\times r$

$mv^2=80\times 2\times 10^7$  

$mv^2=1.6\times 10^9\ kg-m^2/s^2$

Multiply both sides of above equation by 1/2.

$\dfrac{1}{2}mv^2=\dfrac{1}{2}\times 1.6\times 10^9\ kg-m^2/s^2$        

$\dfrac{1}{2}mv^2=8\times 10^8\ J$

So, $E=8\times 10^8\ J$'        

(b) Let F' is the force if the orbit radius were increased to, $r'=3\times 10^7\ m$

The relation between the force and distance is,

$\dfrac{F'}{F}=(\dfrac{r}{r'})^2$

$\dfrac{F'}{80}=(\dfrac{2\times 10^7}{3\times 10^7})^2$        

F' = 35.55 N

or

F' = 36 N

Hence, this is the required solution.