An 8000 kg satellite is launched from the surface of the earth,and injected into a circular orbit at an altitude of 100 km above the surface of the earth.The gravitational potential energy of the satellite in the circular orbit is (G = 6.67 $\times$ 10^-11 Nm²/kg², Me = 5.98 $\times$ 10²⁴ kg, Re = 6.38 $\times$ 10⁶ m) A) 5.01 $\times$ 10¹¹ J. B) 4.92 $\times$ 10¹¹ J. C) 4.75 $\times$ 10¹¹ J. D) 4.02 $\times$ 10¹¹ J. E) 3.85 $\times$ 10¹¹ J.

Question

Quizplus · Accepted Answer

The gravitational potential energy of an object in circular orbit is given by the formula: U = -GMm/2r
where G is the gravitational constant, M is the mass of the Earth, m is the mass of the satellite, and r is the radius of the orbit (which is the sum of the radius of the Earth and the altitude of the orbit).
Plugging in the values, we get:
U = -[(6.67e-11 Nm^2/kg^2)×(5.98e24 kg)×(8000 kg)]/[2×(6.38e6 m + 1e5 m)]
U ≈ -4.92×10^11 J
The negative sign indicates that the potential energy is bound (i.e., the satellite is gravitationally attracted to the Earth). However, we are only interested in the magnitude of the energy, which is 4.92×10^11 J (option B).

An 8000 Kg Satellite Is Launched from the Surface of the Earth,and