Question

Two identical satellites orbit the earth in stable orbits. Onesatellite orbits with a speed vat a distance rfrom the center of the earth. The second satellite travels at aspeed that is less than v.At what distance from the center of the earth does the secondsatellite orbit?At a distance that is less than r.At a distance equal to r.At a distance greater than r.Now assume that a satellite of mass m is orbiting the earth at a distance r from the center of the earth with speed v_e. An identical satellite is orbiting the moon at thesame distance with a speed v_m. How does the time T_m it takes the satellite circling the moon to make onerevolution compare to the time T_e it takes the satellite orbiting the earth to make onerevolution?T_m is less than T_e.T_m is equal to T_e.T_m is greater than T_e.

Answer 1

a. At a distance greater than r

b. T_m is greater than T_e.

Step-by-step explanation:

a. Two identical satellites orbit the earth in stable orbits. One satellite orbits with a speed vat a distance r from the center of the earth. The second satellite travels at a speed that is less than v. At what distance from the center of the earth does the second satellite orbit?

Since the centripetal force on any satellite, F equals the gravitational force F' at r,

and F = mv²/r and F' = GMm/r² where m = mass of satellite, v = speed of satellite, G = universal gravitational constant, M = mass of earth and r = distance of satellite from center of earth.

Now, F = F'

mv²/r = GMm/r²

v² = GM/r

v = √GM/r

Since G and M are constant,

v ∝ 1/√r

So, if the speed decreases, the radius of the orbit increases.

Since the second satellite travels at a speed less than v, its radius, r increases since v ∝ 1/√r.

So, the distance the second satellite orbits is at a distance greater than r

b. An identical satellite is orbiting the moon at the same distance with a speed v_m. How does the time T_m it takes the satellite circling the moon to make one revolution compare to the time T_e it takes the satellite orbiting the earth to make one revolution?

Since the speed of the satellite, v = √GM/r where M = mass of planet

Since the satellite is orbiting at the same distance, r is constant

So, v ∝ √M

Since mass of earth M' is greater than mass of moon, M", the speed of satellite circling moon, v_m is less than v the speed of satellite circling earth at the same distance, r

Now, period T = 2πr/v where r = radius of orbit and v = speed of satellite

Since r is constant for both orbits, T ∝ 1/v

Now, since the speed of the speed of the satellite on earth orbit v is greater than the speed of the satellite orbiting the moon, v_m, and T ∝ 1/v, it implies that the period of the satellite orbiting the earth, T_e is less than the period of the satellite orbiting the moon, T_m since there is an inverse relationship between T and v. T_e is less T_m implies T_m is greater than T_e

So, T_m is greater than T_e.