**Solve the following problem.**

What is the gravitational potential due to the Earth at a point which is at a height of 2R_{E} above the surface of the Earth?

(Mass of the Earth is 6 × 10^{24} kg, radius of the Earth = 6400 km and G = 6.67 × 10^{–11} N m^{2} kg^{–2})

#### Solution

**Given: **M = 6 × 10^{24} kg, R_{E} = 6400 km = 6.4 × 10^{6 }m, G = 6.67 × 10^{–11} N m^{2}/kg^{2}, h = 2R_{E}

**To find:** Gravitational potential (V)

**Formula: **V = -`"GM"/"r"`

**Calculation:** From formula,

V = `- "GM"/("R"_"E" + 2"R"_"E")`

`= - (6.67 xx 10^-11 xx 6 xx 10^24)/(3 xx 6.4 xx 10^6)`

`= -(6.67 xx 2)/6.4 xx 10^7`

= **- 2.08 × 10 ^{7} J kg^{-1}**

Negative sign indicates the attractive nature of gravitational potential.

Gravitational potential due to Earth will be **2.08 × 10 ^{7} J kg^{-1 }**towards the centre of the Earth.