The equation 4x2 T = gives Kepler's third law in a form suitable for objects orbiting the Sun. To modify it to a form suitable for objects in the Earth's orbit, we replace the mass of the Sun with the mass of the Earth to obtain the following expression for the period T of the satellite, where G is the constant of universal gravitation, M, is the mass of the Earth, and r is the radius of the satellite's orbit that we found above. T2 = GME Solving for the period of the satellite, we have T = GMĘ 4x2 x 106 m) (6.67 x 10-11 N - m2/kg2)(1 k9 - m/s5.98 x 1024 kg) 1 N x 103 s. Converting from seconds to hours, we find that the period of the satellite is 1h T: x 103 h. 3600 A satellite is in a circular orbit around the Earth at an altitude of 3.29 x 10° m. (b) Find the speed of the satellite.

Principles of Physics: A Calculus-Based Text
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Chapter11: Gravity, Planetary Orbits, And The Hydrogen Atom
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Please fill in all of the steps to answer part B
The equation
4x2
T =
gives Kepler's third law in a form suitable for objects orbiting the Sun. To modify it to a form suitable for objects in the Earth's orbit, we replace the mass of the Sun with the
mass of the Earth to obtain the following expression for the period T of the satellite, where G is the constant of universal gravitation, M, is the mass of the Earth, and r is the
radius of the satellite's orbit that we found above.
T2 =
GME
Solving for the period of the satellite, we have
T =
GMĘ
4x2
x 106 m)
(6.67 x 10-11 N - m2/kg2)(1 k9 - m/s5.98 x 1024 kg)
1 N
x 103 s.
Converting from seconds to hours, we find that the period of the satellite is
1h
T:
x 103
h.
3600
Transcribed Image Text:The equation 4x2 T = gives Kepler's third law in a form suitable for objects orbiting the Sun. To modify it to a form suitable for objects in the Earth's orbit, we replace the mass of the Sun with the mass of the Earth to obtain the following expression for the period T of the satellite, where G is the constant of universal gravitation, M, is the mass of the Earth, and r is the radius of the satellite's orbit that we found above. T2 = GME Solving for the period of the satellite, we have T = GMĘ 4x2 x 106 m) (6.67 x 10-11 N - m2/kg2)(1 k9 - m/s5.98 x 1024 kg) 1 N x 103 s. Converting from seconds to hours, we find that the period of the satellite is 1h T: x 103 h. 3600
A satellite is in a circular orbit around the Earth at an altitude of 3.29 x 10° m.
(b) Find the speed of the satellite.
Transcribed Image Text:A satellite is in a circular orbit around the Earth at an altitude of 3.29 x 10° m. (b) Find the speed of the satellite.
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