Question

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 780 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

Answer 1

Answer:

a) 3h 59’  b) 5,590 m/s  c) 1,910 N

Explanation:

The only force acting on the satellite (neglecting the influence of any other body) is just the attractive force from Earth, which is given by the Universal Law of Gravitation:

Fg = G* ms*me / (res² (1)

where:  G = 6.67* 10⁻¹¹ N.m² / kg² ms = 780 kg, me= 5.97* 10²⁴ kg, and  

res = 2* 6.38* 10⁶ m (distance between the center of the Earth and the satellite).

Fg = 1,910 N (answer c)

At the same time, this force, is the centripetal force that keeps the satellite in orbit, and that can be written as follows:

Fg = ms * v² / res (2)

By definition of velocity, we can say that the constant speed at which the satellite orbits, can be expressed as the quotient between the distance travelled around the Earth once, and the time needed to do that, which is called the period of the orbit:

v = 2*π*res / T (3)

We can solve for v first, taking the right sides of (1) and (2), as follows:

G* ms*me / (res)²= ms * v² / res

v = √(G*me/r) = 5,590 m/s (answer b)

Once obtained the value of v, we can replace in (3), and solve for T:

T = 2* π*res / v = 3 h 59 min (answer a)