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].
