The Hubble Space Telescope has a mass of 1.16*10^ 4 kg and orbits the Earth at an altitude of 5.68 * 10 ^ 5 above Earth's surface. 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

Answers

Answer 1

Answer:

[tex]E=8.13\times 10^{12}\ J[/tex]

Explanation:

Given that,

The mass of a Hubble Space Telescope, [tex]m_1=1.16\times 10^4\ kg[/tex]

It orbits the Earth at an altitude of [tex]5.68\times 10^5\ m[/tex]

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

[tex]E=\dfrac{Gm_1m_e}{r}[/tex]

Where

[tex]m_e[/tex] is the mass of Earth

Put all the values,

[tex]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[/tex]

So, the potential energy of the telescope is [tex]8.13\times 10^{12}\ J[/tex].


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