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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

Sagot :

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Answer:

[tex]-5.00\times 10^{12}\ \text{J}[/tex]

Explanation:

m = Mass of satellite = 175000 kg

h = Distance above Earth = 600 km

R = Radius of Earth = 6371 km

M = Mass of Earth = [tex]5.972\times 10^{24}\ \text{kg}[/tex]

G = Gravitational constant = [tex]6.674\times 10^{-11}\ \text{Nm}^2/\text{kg}^2[/tex]

Total energy is given by

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

The total energy of the satellite is [tex]-5.00\times 10^{12}\ \text{J}[/tex].

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