A)What is the speed of a 250 kg satellite in an approximately circular orbit 600

Question

A)What is the speed of a 250 kg satellite in an approximately circular orbit 600 km above the surface of Earth? B)What is the period of a 250 kg satellite in an approximately circular orbit 600 km above the surface of Earth? C)Suppose the satellite losses mechanical energy at the average rate of 2.86×104 J per orbital revolution. Adopting the reasonable approximation that the satellite's orbit becomes a "circle of slowly diminishing radius," determine the satellite's altitude at the end of its 1500th revolution. D)And determine the satellite's speed at the end of its 1500th revolution. E)And determine the satellite's period at the end of its 1500th revolution. F)What is the magnitude of the average retarding force on the satellite? (Use the value for the first orbit, enter as a positive number)

Explanation / Answer

a) Assuming that the mass of Earth is 5.97x1024 kg and the radius of Earth is 6.37x106 m the expression to use is:

mv2/r = GMm/r2
v2 = GM/r
v = (GM/r)1/2
G = 6.673x10-11 m3/kg s2
r = 6.37x106 + 600000 = 6.97x106 m

v = (6.673x10-11 * 5.97x1024 / 6.97x106)1/2
v = 7560.16 m/s

b) For the period, Use the following expression:
T = (4pi2r3 / GM)1/2
T = (4*(3.1416)2(6.97x106)3 / 6.673x10-11 * 5.97x1024)1/2
T = 5792.7 s

c) E = Total Energy Lost = energy lost per turn x 1500 revs
= 2.86x104 * 1500 = 4.29x107 J

Now use the following expression:
E1 = -GMm/2r
E1 = -6.673x10-11 * 5.97x1024 * 250 / 2*6.97x106 = -7.14x109 J

E2 = -7.14x109 - 4.29x107 = -7.183x109 J

E2 = -GMm/2r ---> solve for r
r = 6.673x10-11 * 5.97x1024 * 250 / 2*7.183x109
r = 6.933x106 m

Then the distance:
6.933x106 - 6.37x106 = 5.623x105 m

d) Now that we know the value of r, we use this value to get speed and period as in part a and b:
v = (6.673x10-11 * 5.97x1024 / 6.933x106)1/2
v = 7580.31 m/s

e) For the period:
T = 2pir/v
T = 2*3.1416*6.933x106 / 7580.31
T = 5746.65 s

f) For the magnitude:
F = E/2pir
F = 2.86x104 / 2*3.1416*6.933x106
F = 6.5654x10-4 N

Hope this helps

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