Question

A 1,160 kg satellite orbits Earth with a tangential speed of 7,446 m/s. If the satellite experiences a centripetal force of 8,955 N, what is the height of the satellite above the surface of Earth? Recall that Earth’s radius is 6.38 × 106 m and Earth’s mass is 5.97 × 1024 kg.

Answer 1

Answer:

$h = 8.14 \times 10^5 m$

Explanation:

Let the height above the surface of Earth is H

now we know that gravitational force due to Earth on the satellite is net centripetal force due to which it is revolving around the Earth

So we can say

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

$\frac{GM m}{(R + h)^2} = \frac{mv^2}{(R + h)}$

now we know that

$\frac{GM}{(R + h)} = v^2$

$\frac{(6.67 \times 10^{-11})(5.98 \times 10^{24})}{(6.38 \times 10^6 + h)} = 7446^2$

$h = 8.14 \times 10^5 m$

Answer 2

Answer: The height of the satellite above the surface of Earth is $8.02\times 10^{5}m$

Explanation:

Given

Mass of the satellite, m= 1160 kg

tangential speed , v = 7446 m/s

Centripetal force , F = 8955 N

Radius of earth , R= $6.38\times 10^{6}$ m

Let height of satellite above surface of the Earth be H

Centripetal force on satellite is given by

$F=\frac{mv^{2}}{R+H}$

=>$H=\frac{mv^{2}}{F}-R$

=>$H=(\frac{1160\times 7446^{2}}{8955}-6.38\times 10^{6}) m=8.02\times 10^{5}m$

Thus the height of the satellite above the surface of Earth is $8.02\times 10^{5}m$