Question

# NASA launches a satellite into orbit at a height above the surface of the Earth equal...

NASA launches a satellite into orbit at a height above the surface of the Earth equal to the Earth's mean radius. The mass of the satellite is 830 kg. (Assume the Earth's mass is 5.97 1024 kg and its radius is 6.38 106 m.) (a) How long, in hours, does it take the satellite to go around the Earth once? h (b) What is the orbital speed, in m/s, of the satellite? m/s (c) How much gravitational force, in N, does the satellite experience? N

then part B ) For orbit,

Centripetal acceleration = Gravitational acceleration

v^2/r = GM/r^2

v = sqrt(GM/r)

where, G = Newton's gravitational constant = 6.674*10^-11 N m^2 kg^-2

M = mass of central body = 5.97*10^24 kg

r = orbit radius = 2 * 6.38*10^6 m = 12.76*10^6 m

v = sqrt(6.674*10^-11*5.97*10^24/(12.76*10^6))

= 5587.98 m/s

part B) Circumference = 2pi(2r)

= 2*pi*2*638*10^6 m

= 8.02*10^7 m

Time t = Circumference / Velocity

= 8.02*10^7 m / 5587.98 m/s

= 3.9867 h

part C) Easie is to use the inverse-square-of-distance law

F = mg(r / 2r)^2

= 830 kg * 9.8 m/s^2/4

= 2033.5 N

#### Earn Coins

Coins can be redeemed for fabulous gifts.