A 745-kg satellite is in a circular orbit about Earth at a height above Earth eq

Question

A 745-kg satellite is in a circular orbit about Earth at a height above Earth equal to Earth's mean radius. (a) Find the satellite's orbital speed. 7913.047 Changed: Your submitted answer was incorrect. Your current answer has not been submitted. Your response differs from the correct answer by more than 10%. Double check your calculations. m/s (b) Find the period of its revolution. h (c) Find the gravitational force acting on it. N I don't know what I've done wrong on part a. Can someone please check my answer and show me the work on a, b, and c? Thank you in advance.

Explanation / Answer

M = Mass of Earth
G = Universal Gravitational Constant
Re = Radius of Earth
Sa = Satellite Altitude
R = Total Radius

Values are:

M = 5.9736E+24 kg
G = 6.67428E-11 m^3/kg-s^2
Re = 6.371E+6 m
Sa = 6.371E+6 m
R = 1.2742E+7 m

(a) Orbital velocity

v = SQRT { [GM] / R }
v = SQRT { [ (6.67428E-11 m^3/kg-s^2) * (5.9736E+24 kg) ] / (1.2742E+7 m) }
v = SQRT { [ 3.9869E+14 m^3/s^3 ] / (1.2742E+7 m) }
v = SQRT { 31,289,812 m^2/s^2 }
v = 5,594 m/s

(b) Orbital period

T = SQRT { [4 * ^2 * R^3] / [GM] }
T = SQRT { [4 * ^2 * (1.2742E+7 m)^3 ] / [ (6.67428E-11 m^3/kg-s^2) * (5.9736E+24 kg) ] }
T = SQRT { [ (39.478) * (2.0688E+21 m^3) ] / [ 3.9869E+14 m^3/s^2 ] }
T = SQRT { [ 8.167E+22 m^3 ] / [ 3.9869E+14 m^3/s^2 ] }
T = SQRT { 204,847,772 s^2 }
T = 14,313 s

c) Gravitational force: GMm/r^2
G is newtons constant, 6.673 x 10^-11
M is mass of planet
m is mass of satellite (745 kg)
r is distance of satellite from CENTER of planet
Plug in the numbers: 6.673 x 10^-11 x 5.9742 x 10^24 x 745/12,765,200^2
Gravitational force= 1822.64 Newtons

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