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A satellite is in a circular orbit around the Earth at an altitude of 3.23 x 106 m (a) Find the period of the orbit. (b) Find the speed of the satellite. (c) Find the acceleration of the satellite. Step 1 (a) If the satellite has an altitude of h - 3.23 x 106 m above the surface of the Earth, the radius of its orbit r, where RE is the Earths radius, is given by h(638 10)32332106610 m 9.61 106 m rR th-6.38 x 10° m +(3.23 3.23 x 10 m9.61The equation 2 4J0 GM gives Keplers third law in a form suitable for objects orbiting the Sun. To modify it to a form suitable for objects in the Earths 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 satellites orbit that we found above 47t 2 GM Solving for the period of the satellite, we have 472r3 GM 4T2 9.61 x 106 m (667-10-11N m2p 2.98 104 k) x 10-11 N m2/kg2 Your response differs significantly from the correct answer. Rework your solution from the beginnin Converting from seconds to hours, we find that the period of the satellite is Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. x 105 s Your response differs from the correct answer by more than 100%. h. 1 h 3

please explain how to get rid of the square and get the answer for C part as well.

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2. GnM 3 2- 878.12 xi s 9332 S 24

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please explain how to get rid of the square and get the answer for C part...
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