Question

Many people assume air resistance acting on a moving object will always make the object slow down. It can, however, actually be responsible for making the object speed up. Consider a 310-kg Earth satellite in a circular orbit at an altitude of 335 km. A small force of air resistance makes the satellite drop into a circular orbit with an altitude of 200 km. (Use the following values: G = 6.67 10-11 m3 kg−1 s−2, mass of the Earth 5.98 1024 kg, radius of the Earth 6.37 106 m.)

(a) Calculate the satellite's initial speed.
_________ m/s

(b) Calculate its final speed in this process.
_________ m/s

(c) Calculate the initial energy of the satellite-Earth system. (Use the exact values you enter in previous answer(s) to make later calculation(s). Round your answer to two decimal places.)
_________×10⁹ J

(d) Calculate the final energy of the system. (Use the exact values you enter in previous answer(s) to make later calculation(s). Round your answer to two decimal places.)
_________ ×10⁹ J

(e) Show that the system has lost mechanical energy.

Find the amount of the loss due to friction. (Round your answer to one decimal place.)
___________ ×10⁸ J

(f) What force makes the satellite's speed increase? Hint: You will find a free-body diagram useful in explaining your answer.

_______________________________

Answer 1

Remember that total radius is from center of the Earth, not surface altitude. Using Kepler's circular mechanics. Do not use circular motion equations! They will give you incorrect answers, as they do not apply to orbits. Given values:

Me = 5.98E+24 kg
Re = 6,370,000 m
G = 6.67E-11 m^3/kg-s^2
m = 310 kg
Ri = 6705000 m
Rf = 6,570,000 m
g = 9.81 m/s^2

Where:

Me = Mass of the Earth
Re = Radius of the Earth
G = Universal Gravitational Constant
m = Mass of satellite
Ri = Initial orbital radius
Rf = Final orbital radius
g = Earth gravity

(a) Initial orbital velocity

v = SQRT { [GM] / Ri }
v = SQRT { [ (6.67E-11 m^3/kg-s^2) * (5.98E+24 kg) ] / (6705000 m) }
v = SQRT { [ 3.98866E+14 m^3/s^2 ] / (6705000 m) }
v = SQRT { 59487844.89 m^2/s^2 }
v = 7712.84 m/s

(b) Final orbital velocity

v = SQRT { [GM] / Rf }
v = SQRT { [ (6.67E-11 m^3/kg-s^2) * (5.98E+24 kg) ] / (6,570,000 m) }
v = SQRT { [ 3.98866E+14 m^3/s^2 ] / (6570000 m) }
v = SQRT {60710197.87 m^2/s^2 }
v = 7791.67 m/s

(c) Initial orbital energy

KE = 0.5 * m * v^2
KE = 0.5 * (310 kg) * (7712.84 m/s)^2
KE = (155 kg) * (59487900.86 m^2/s^2)
KE = 9220624634.2 J

(d) Final orbital energy

KE = 0.5 * m * v^2
KE = 0.5 * (310 kg) * (7791.67 m/s)^2
KE = (155 kg) * (60710121.4 m^2/s^2)
KE = 9410068815.2795 J

(e) Mechanical energy loss

Initial -->
ME = PE + KE
ME = [m*g*Ri] + [ 0.5 * m * v^2 ]
ME = [ (310 kg) * (9.81 m/s^2) * (6705000 m) ] + [ 0.5 * (310 kg) * (7712.84 m/s)^2 ]
ME = [ 20390575500 J ] + [ (155 kg) * (59487900.9 m^2/s^2) ]
ME = [20390575500 J ] + [ 9220624634.168 J ]
ME = 29611200134.2 J

Final -->
ME = PE + KE
ME = [m*g*Rf] + [ 0.5 * m * v^2 ]
ME = [ (310 kg) * (9.81 m/s^2) * (6570000 m) ] + [ 0.5 * (310 kg) * (7791.67 m/s)^2 ]
ME = [ 19980027000J ] + [ (155 kg) * (60710121.4 m^2/s^2) ]
ME = [ 19980027000 J ] + [ 9410068815.3 J ]
ME = 29390095815.3 J

Loss = Initial - Final
Loss = (29611200134.2 J) - (29390095815.3 J)
Loss = 221104318.9 J

(f) Force that increases speed

Initial gravitational -->
a = GM / r^2
a = [GM] / Ri
a = [ (6.67E-11 m^3/kg-s^2) * (5.98E+24 kg) ] / (6705000 m)^2
a = [ 3.98866E+14 m^3/s^2 ] / (4.496E+13 m^2)
a = 8.87 m/s^2

Final gravitational -->
a = GM / r^2
a = [GM] / Rf
a = [ (6.67E-11 m^3/kg-s^2) * (5.98E+24 kg) ] / (6570000 m)^2
a = [ 3.98866E+14 m^3/s^2 ] / (4.316E+13 m^2)
a = 9.24 m/s^2

Initial force -->
F = m * a
F = (310 kg) * (8.87 m/s^2)
F = 2749.7 N

Final force -->
F = m * a
F = (310 kg) * (9.24 m/s^2)
F = 2864.4 N

The force on the satellite due to Earth gravity increased by 114.7 N through its fall, during which time the satellite was under uniform acceleration due to the changing radius, thus increasing its speed.