A remote sensing satellite of the earth revolves in a circular orbit at a height of 250 km above the earth’s surface. What is the orbital speed and period of revolution of the satellite?

Solution

Given,

Height of satellite (h) = 250 km = 250000 m

Orbital speed (v) =?

Period of revolution of the satellite (T) =?

Radius of earth (r) = [latex]6.36 \times 10^6[/latex] m

Mass of earth (M) = [latex]6 \times 10^{24}[/latex] kg

We know that,

[latex]v = \sqrt{\dfrac{GM}{r+h}}[/latex]

[latex]= \sqrt{\dfrac{(6.67 \times 10^{-11}) \times (6 \times 10^{24})}{(6.36 \times 10^6 + 250000)}}[/latex]

[latex]= 7.781 \times 10^3\ m/s[/latex]

Now, the force of attraction between earth and satellite provides necessary centripetal force. So,

Centripetal force = Gravitational force