Question

Consider a satellite of mass ms in circular orbit around Earth, a distance h above Earth's surface. Assume the Earth is a sphere with radius Re and mass Me.

(a) As the satellite travels in circular orbit, will its speed increase, decrease, or remain constant? Explain.

(b) The only force acting on the satellite is gravity, so the satellite is in freefall. Why doesn't the satellite get closer to Earth's surface?

(c) Determine the ratio of the force of gravity on the satellite while in orbit to the force of gravity on the satellite while on the surface of Earth. Simplify your answer; no complex fractions! Give your answer in terms of Re and h.

(d) Starting with Newton's second law, determine an equation for each of the following quantities. Give your answers in terms of Me, Re, h, and the universal gravitation constant G.

(i) the freefall acceleration of the satellite,

(ii) The orbital speed of the satellite,

(iii) the period of orbit, which is the time it takes the satellite to complete one orbit around Earth.

(e) Determine the work done by gravity as the satellite moves through one-half of its circular orbit.

(f) Consider a near-Earth orbiting satellite (h=0.10Re) that is inhabited by astronauts.

(i) Determine the ration of the freefall acceleration of an astronaut in this circular orbit to the freefall acceleration of an astronaut on Earth.

(ii) If that astronaut weights 150 pounds on Earth, how large is the force of gravity on her in orbit? Is she "weightless"?

Answer 1

a) In a circular orbit, speed always remains constant. The change is speed is observed in elliptical orbits. In circular orbit, distance is always constant and hence speed, which is in accordance with the law of constant aerial velocity.

b) It is true that the atellites are in freefall, but the rate at which the falls toward earth is eqaul to the rate at which earth curves away from satellite. It is similar to a situation of a dog trying to catch its own tail.

c) Force on surface of earth Fs= (G*Me*Ms)/Re^2

Force in orbit Fo= (G*Me*Ms)/(Re+h)^2

Fo/Fs = [Re/(Re+h)]^2

d)i) F = G*Me*Ms / (Re + h)^2 = Ms*a

a = G*Me / (Re + h)^2

ii) a = V^2 / (Re + h)

V = sqrt[ a*(Re+h)]

V = sqrt [G*Me / (Re+h)]

iii) T = 2*pi*(Re+h) / V

T = [2*pi*(Re+h)] / sqrt [G*Me / (Re+h)]

T = [2*pi*(Re+h)^(3/2)] / sqrt(G*Me)

e) As the force of gravity is directed toward centre of earth and displacement is in perpendicular direction, hence,

work done by gravity is zero

f)i) acceleration in orbit = G*Me / (Re + 0.1Re)^2 = G*Me / (1.21*Re^2)

acceleration on surface = G*Me / Re^2

Ratio = [G*Me / (1.21*Re^2)] / [ G*Me / Re^2]

= 0.826

ii) in the orbit the wieght will reduce by the ratio calculayed in above part i.e

W = 150*0.826 = 123.96 lb