Question

1- If a satellite orbiting the Earth at a height of 150 km (93 miles) above the surface of the earth.
Calculate the speed, acceleration and orbital period of the satellite. Using these given constants:
M(earth) = 5.98 x 10^24 kg, R(earth) = 6.37 x 10^6 m).

Answer 1

F = G m M/r^2
G = 6.67*10^-11
m is mass of our thing
M = 5.98*10^24
r = 6.37*10^6 + 150,000 = 6.37*10^6 + 0.150 *10^6 = 6.52*10^6
so
F = m * 6.67*10^-11 * 5.98*10^24 / 42.5*10^12
= m * 9.38
{{ note - if we were at earth surface it would be about m * 9.81 , we did not go up much :) }}
Now the problem:
F = mass * centripetal a = m * v^2/r
so
9.38 = v^2/r = v^2 / 6.52*10^6
v^2 = 9.38 * 6.52 * 10^6 meters/second
solve for v
then a = v^2/r
T = 2 pi r / v