Question

Q1: A 1 036-kg satellite orbits the Earth at a constant altitude of 93-km.

(a) How much energy must be added to the system to move the satellite into a circular orbit with altitude 203 km?
MJ

(b) What is the change in the system's kinetic energy?
MJ

(c) What is the change in the system's potential energy?
MJ

Q2:

A 475 kg satellite is in a circular orbit at an altitude of 575 km above the Earth's surface. Because of air friction, the satellite eventually falls to the Earth's surface, where it hits the ground with a speed of 2.00 km/s. How much energy was transformed into internal energy by means of air friction?

Q3:

When a falling meteoroid is at a distance above the Earth's surface of 3.10 times the Earth's radius, what is its acceleration due to the Earth's gravitation?

Answer 1

Solution) m = 1036 kg

h1 = 93 km = 93000 m

(a) h2 = 203 km = 203000 m

E = ?

E = - ( GMm)/(2r)

r1 = R + h1

R = 6.37×10^(6) m

r1 = 6.37×10^(6) + 93000 = 6463000 m

r2 = R + h2

r2 = 6.37×10^(6) + 203000

r2 = 6573000 m

E = - (GMm)/(2(r2)) - ( - (GMm)/(2(r1)))

E = (GMm/2)((1/r1) - (1/r2))

E = ((6.67×10^(-11)×5.98×10^(24)×1036)/2)((1/6463000) - (1/6573000))

E = 535 MJ

(b) Change in systems kinetic energy , K = ?

K = - E

K = - 535 MJ

(c) change in systems potential energy , U = ?

U = 2E

U = 2×535 = 1070 MJ

* kindly post next questions in next post