Question

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

Answer 1

`E=8.13* 10^(12)\ J`

Step-by-step explanation:

Given that,

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

It orbits the Earth at an altitude of
`5.68* 10^5\ m`

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

`E=(Gm_1m_e)/(r)`

Where

`m_e` is the mass of Earth

Put all the values,

`E=(6.67* 10^(-11)* 1.16* 10^4* 5.97* 10^(24))/(5.68* 10^5)\\\\E=8.13* 10^(12)\ J`

So, the potential energy of the telescope is
`8.13* 10^(12)\ J`.