Question

A certain satellite is in circular orbit about the earth at an altitude of 550km. If the satellite makes a revolution in 110minutes, calculate its orbital speed ​

Answer 1

Approximately
`6.59* 10^(3)\; \rm m\cdot s^(-1)`.

Step-by-step explanation:

Look up the radius of the earth: approximately
`6.371 * 10^(3)\; \rm km`.

The radius of the orbit of this satellite is the of the radius of the earth (at ground level) plus the height of the satellite relative to ground level:

`\begin{aligned}r &\approx 6.371 * 10^(3)\; {\rm km} + 550\; {\rm km} \\ &= 6.921 * 10^(3)\; \rm km\end{aligned}`.

Convert the units of both distance and time to standard units:

Orbital radius:

`\begin{aligned}r &\approx 6.921 * 10^(3)\; {\rm km} \\ &= 6.921 * 10^(3)\; {\rm km} * (10^(3)\; \rm m)/(1\; \rm km) \\ &= 6.921 * 10^(6)\; \rm m\end{aligned}`.

Orbital period:

`\begin{aligned}t &= 110\; \text{minute} \\ &= 110\; \text{minute} * \frac{60\; \text{second}}{1\; \text{minute}} \\ &= 6.6 * 10^(3)\; \text{second}\end{aligned}`.

Calculate the circumference of this orbit:

`\begin{aligned}& 2\, \pi\, r \\ \approx\; & 2 \, \pi * 6.921 * 10^(6)\; {\rm m} \\ \approx\; & 4.35 * 10^(7)\; \rm m\end{aligned}`.

Calculate the orbital speed (tangential) of this satellite:

`\begin{aligned}v &= (2\, \pi\, r)/(t) \\ &\approx (4.35 * 10^(7)\; \rm m)/(6.6 * 10^(3)\; \rm s) \\ &\approx 6.6 * 10^(3)\; \rm m \cdot s^(-1)\end{aligned}`.