Question

The radius of the Earth is 6370 km. If a satellite orbits 150 km above the Earth's surface, what is the velocity of the orbit? The Earth's mass is 5.97 × 1024 kg. Give your answer in m-s¹, to three significant figures and do not include units.​

Answer 1

Approximately
`7.61 * 10^(3)\; {\rm m\cdot s^(-1)}`.

Step-by-step explanation:

Add the radius of the Earth to the altitude of the satellite to find the radius of the orbit:

`\begin{aligned}r &= (6370 + 150)\; {\rm km} \\ &= 6880\; {\rm km} \\ &= 6.88* 10^(6)\; {\rm m}\end{aligned}`.

Look up the gravitational constant:

`G \approx 6.6743* 10^(-11)\; {\rm m^(3)\, kg^(-1)\, s^(-2)`.

Let
`m` denote the mass of the satellite. At an orbital velocity of
`v`, the (centripetal) net force on the satellite would be:

`\displaystyle (\text{net force}) = (m\, v^(2))/(r)`.

Let
`M` denote the mass of planet Earth. At a distance of
`r` from the center of the Earth, the gravitational attraction on the satellite would be:

`\displaystyle (\text{gravitational attraction}) = (G\, M\, m)/(r^(2))`.

When the satellite is at the correct orbital velocity, the net force on the satellite would be equal to the gravitational attraction from the Earth. In other words:

`\displaystyle (\text{net force}) = (\text{gravitational attraction})`.

`\displaystyle (m\, v^(2))/(r) = (G\, M\, m)/(r^(2))`.

Rearrange the equation and solve for orbital velocity
`v`:

`\begin{aligned}v^(2) &= (G\, M)/(r)\end{aligned}`.

`\begin{aligned}v &= \sqrt{(G\, M)/(r)} \\ &\approx \sqrt{((6.6743 * 10^(-11))\, (5.97 * 10^(24)))/(6.88 * 10^(6))}\; {\rm m\cdot s^(-1)} \\ &\approx 7.61 * 10^(3)\; {\rm m\cdot s^(-1)}\end{aligned}`.