Question

A 40.9 kg satellite circles planet Cruton every 5.82 h. The magnitude of the gravitational force exerted on the satellite by Cruton is 105 N. (a) What is the radius of the orbit

Answer 1

Answer:

r = 2.8524 x10⁷ m = 28524 km

Explanation:

First, we calculate the angular speed of the satellite:

$\omega = \frac{Arc Length}{Time}\\\\For\ One\ Complete\ Revolution\ around\ Cruton:\\\\\omega = \frac{2\pi\ rad}{5.82\ h}\frac{1\ h}{3600\ s}\\\omega = 3\ x\ 10^{-4} rad/s$

Now, we use the formula for the gravitational force. Since gravitational force will be acting as the centrifugal force due to circular motion. Therefore,

$F = \frac{mv^{2}}{r}$

where,

F = Gravitational Forc = 105 N

m = mass of satellite = 40.9 kg

v = linear speed of satellite = rω = r(3 x 10⁻⁴ rad/s)

r = radius of orbit = ?

Therefore,

$105\ N = \frac{(40.9\ kg)(r*3\ x\ 10^{-4}\ rad/s)^{2}}{r}\\r = \frac{105}{(40.9)(9\ x\ 10^{-8})}$

r = 2.8524 x10⁷ m = 28524 km