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3 years ago

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The Hubble Space Telescope has a mass of 1.1610^ 4 kg and orbits the Earth at an altitude of 5.68 10 ^ 5 above Earth's surfac

e. Relative to infinitydetermine the potential energy the telescope at this location. Would the formula be Ep=-Gm1m2/r or positive G since it’s relative to infinity

1 answer:

andrezito [222]3 years ago

5 0

Answer:

$E=8.13\times 10^{12}\ J$

Explanation:

Given that,

The mass of a Hubble Space Telescope, $m_1=1.16\times 10^4\ kg$

It orbits the Earth at an altitude of $5.68\times 10^5\ m$

We need to find the potential energy the telescope at this location. The formula for potential energy is given by :

$E=\dfrac{Gm_1m_e}{r}$

Where

$m_e$ is the mass of Earth

Put all the values,

$E=\dfrac{6.67\times 10^{-11}\times 1.16\times 10^4\times 5.97\times 10^{24}}{5.68\times 10^5}\\\\E=8.13\times 10^{12}\ J$

So, the potential energy of the telescope is $8.13\times 10^{12}\ J$ .

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