Answer:
a) 1.24 h
b) [tex]5.132 \times 10^{9} J[/tex]
Explanation:
a) The gravitational force of attraction between satellite and the planet,
which provides centripetal force to keep satellite in circular orbit
[tex]\frac{G M_{p} m}{(R+h)^{2}} &=\frac{m v^{2}}{(R+h)}[/tex]
[tex]v =\sqrt{\frac{G M_{p}}{R+h}}[/tex]
[tex]=\sqrt{\frac{\left(6.67 \times 10^{-11} N \cdot m ^{2} / kg ^{2}\right)\left(1.3 \times 10^{25} kg \right)}{7.2 \times 10^{6} m +400 \times 10^{3} m }}[/tex]
[tex]=1.068 \times 10^{4} m / s[/tex] .......(i)
The expression for time period of the satellite is,
[tex]T=\frac{2 \pi(R+h)}{v}[/tex]
[tex]=\frac{2 \pi\left(7.2 \times 10^{6} m +400 \times 10^{3} m \right)}{1.068 \times 10^{4} m / s }[/tex]
[tex]=4468.9 s \left(\frac{1.0 h }{3600 s }\right)[/tex]
[tex]=1.24 h[/tex]
b) We have from (i) velocity of the satellite as
[tex]v=1.068 \times 10^{4} m / s[/tex]
Kinetic energy of the satellite is,
[tex]K E &=\frac{1}{2} m v^{2}[/tex]
[tex]=\frac{1}{2}(90 kg )\left(1.068 \times 10^{4} m / s \right)^{2}[/tex]
[tex]=5.132 \times 10^{9} J[/tex]
