Question

A 95.0 kg satellite moves on a circular orbit around the Earth at the altitude h=1.20*103 km. Find: a) the gravitational force exerted by the Earth on the satellite? b) the centripetal acceleration of the satellite? c) the speed of the satellite? d) the period of the satellite’s rotation around the Earth?

Answer 1

Answer:

a) $F = 660.576\,N$, b) $a_{c} = 6.953\,\frac{m}{s^{2}}$, c) $v \approx 7255.423\,\frac{m}{s}$, $\omega = 9.583\times 10^{-4}\,\frac{rad}{s}$, d) $T \approx 1.821\,h$

Explanation:

a) The gravitational force exerted by the Earth on the satellite is:

$F = G\cdot \frac{m\cdot M}{r^{2}}$

$F = \left(6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}} \right)\cdot \frac{(95\,kg)\cdot (5.972\times 10^{24}\,kg)}{(7.571\times 10^{6}\,m)^{2}}$

$F = 660.576\,N$

b) The centripetal acceleration of the satellite is:

$a_{c} = \frac{660.576\,N}{95\,kg}$

$a_{c} = 6.953\,\frac{m}{s^{2}}$

c) The speed of the satellite is:

$v = \sqrt{a_{c}\cdot R}$

$v = \sqrt{\left(6.953\,\frac{m}{s^{2}} \right)\cdot (7.571\times 10^{6}\,m)}$

$v \approx 7255.423\,\frac{m}{s}$

Likewise, the angular speed is:

$\omega = \frac{7255.423\,\frac{m}{s} }{7.571\times 10^{6}\,m}$

$\omega = 9.583\times 10^{-4}\,\frac{rad}{s}$

d) The period of the satellite's rotation around the Earth is:

$T = \frac{2\pi}{\left(9.583\times 10^{-4}\,\frac{rad}{s} \right)} \cdot \left(\frac{1\,hour}{3600\,s} \right)$

$T \approx 1.821\,h$