Question

What is the total energy of a 175,000 kg shuttle orbiting the Earth at 600 km above Earth’s surface? (Choose the closest answer)

A -2.46 x 1012

B -3.46 x 1012

C -8.46 x 1012

D -5.46 x 1012

Answer 1

`-5.00* 10^(12)\ \text{J}`

Step-by-step explanation:

m = Mass of satellite = 175000 kg

h = Distance above Earth = 600 km

R = Radius of Earth = 6371 km

M = Mass of Earth =
`5.972* 10^(24)\ \text{kg}`

G = Gravitational constant =
`6.674* 10^(-11)\ \text{Nm}^2/\text{kg}^2`

Total energy is given by

`E=-(GMm)/(2(R+h))\\\Rightarrow E=-(6.674* 10^(-11)* 5.972* 10^(24)* 175000)/(2(6371+600)* 10^3)\\\Rightarrow E=-5.00* 10^(12)\ \text{J}`

The total energy of the satellite is
`-5.00* 10^(12)\ \text{J}`.