LEO orbit

Problem

A satellite follows a circular orbit at altitude $h = 300\;\mathrm{km}$ above the Earth’s surface. Given $g_0 = 9{,}81\;\mathrm{m/s^2}$, $R_T = 6{,}37\cdot 10^6\;\mathrm{m}$, and that the gravitational acceleration at distance $r$ from the centre is $g(r) = g_0\,(R_T/r)^2$, determine the orbital speed.

Gravity (directed towards the centre of the Earth) provides the centripetal force that keeps the satellite in circular orbit.

Solution

The orbital radius, measured from the centre of the Earth, is the sum of the Earth’s radius and the altitude: $r = R_T + h = 6{,}37\cdot 10^6 + 3{,}00\cdot 10^5 = 6{,}67\cdot 10^6\;\mathrm{m}$ The gravitational acceleration at that distance: $g(r) = g_0\left(\frac{R_T}{r}\right)^2 = 9{,}81 \cdot \left(\frac{6{,}37\cdot 10^6}{6{,}67\cdot 10^6}\right)^2 = 9{,}81 \cdot (0{,}955)^2 \approx 8{,}95\;\mathrm{m/s^2}$ In circular orbit gravity is the only force and coincides with the centripetal force, so the gravitational acceleration is the centripetal acceleration: $g(r) = \frac{v^2}{r} \quad\Longrightarrow\quad v = \sqrt{g(r)\cdot r}$ Substituting: $v = \sqrt{8{,}95 \cdot 6{,}67\cdot 10^6} = \sqrt{5{,}97\cdot 10^7} \approx 7{,}72\cdot 10^3\;\mathrm{m/s}$ $\ev{v = \sqrt{g(r)\,r} \approx 7{,}72\;\mathrm{km/s}}$

Linked items

Topics: Dinamica Concepts: Legge di gravitazione universale · Forza centripeta · Accelerazione centripeta · Moto circolare uniforme Skills: Applicazione delle leggi di Newton Objects: Satellite