1.) 7382 m/s
The gravitational attraction between the satellite and the Earth provides the centripetal force that keeps the satellite in circular motion, so we can write
[tex]\frac{GMm}{(R+h)^2}=m\frac{v^2}{R+h}[/tex]
where
G is the gravitational constant
[tex]M=5.98\cdot 10^{24}kg[/tex] is the Earth's mass
m is the satellite's mass
[tex]R=6370 km = 6.37\cdot 10^6 m[/tex] is the Earth's radius
[tex]h=950 km = 0.95\cdot 10^6 m[/tex] is the altitude of the satellite above the Earth's surface
v is the orbital speed
Solving the formula for v, we find
[tex]v=\sqrt{\frac{GM}{R+h}}=\sqrt{\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24} kg)}{(6.37\cdot 10^6 m +0.95\cdot 10^6 m)}}=7382 m/s[/tex]
2) 1.73 hours
The period of the orbit is the time taken to complete one revolution around the Earth, therefore:
[tex]T=\frac{2\pi (R+h)}{v}[/tex]
where the numerator is the circumference of the orbit and v the orbital speed, therefore we find
[tex]T=\frac{2\pi (6.37\cdot 10^6 m+0.95e6 m)}{7382 m/s}=6227 s[/tex]
Converting into hours,
[tex]T=\frac{6227 s}{3600 s/h}=1.73 h[/tex]
