Answer:
a) [tex]F = 660.576\,N[/tex], b) [tex]a_{c} = 6.953\,\frac{m}{s^{2}}[/tex], c) [tex]v \approx 7255.423\,\frac{m}{s}[/tex], [tex]\omega = 9.583\times 10^{-4}\,\frac{rad}{s}[/tex], d) [tex]T \approx 1.821\,h[/tex]
Explanation:
a) The gravitational force exerted by the Earth on the satellite is:
[tex]F = G\cdot \frac{m\cdot M}{r^{2}}[/tex]
[tex]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}}[/tex]
[tex]F = 660.576\,N[/tex]
b) The centripetal acceleration of the satellite is:
[tex]a_{c} = \frac{660.576\,N}{95\,kg}[/tex]
[tex]a_{c} = 6.953\,\frac{m}{s^{2}}[/tex]
c) The speed of the satellite is:
[tex]v = \sqrt{a_{c}\cdot R}[/tex]
[tex]v = \sqrt{\left(6.953\,\frac{m}{s^{2}} \right)\cdot (7.571\times 10^{6}\,m)}[/tex]
[tex]v \approx 7255.423\,\frac{m}{s}[/tex]
Likewise, the angular speed is:
[tex]\omega = \frac{7255.423\,\frac{m}{s} }{7.571\times 10^{6}\,m}[/tex]
[tex]\omega = 9.583\times 10^{-4}\,\frac{rad}{s}[/tex]
d) The period of the satellite's rotation around the Earth is:
[tex]T = \frac{2\pi}{\left(9.583\times 10^{-4}\,\frac{rad}{s} \right)} \cdot \left(\frac{1\,hour}{3600\,s} \right)[/tex]
[tex]T \approx 1.821\,h[/tex]
