Question

NASA wants a satellite to revolve around Earth 3 times a day. What should be the radius of its orbit if we neglect the presence of the Moon? (G = 6.67 × 10-11 N ∙ m2/kg2, Mearth = 5.97 × 1024 kg)

Answer 1

Approximately
`2.03 * 10^(7)\; {\rm m}`.

Step-by-step explanation:

Assume that the radius of this orbit is
`r`.

Let
`m` denote the mass of this satellite and let
`M` denote the mass of the Earth. At a distance of
`r` from the center of the earth, the magnitude of the gravitational attraction on this satellite would be
`G\, m\, M / (r^(2))`.

The question implies that the gravitational pull from the earth is the only significant force on this satellite. Hence, the net force on this satellite would be also
`G\, m\, M / (r^(2))`.

The acceleration of this satellite would thus be
`a = (\text{net force}) / (\text{mass}) = G\, M / (r^(2))`.

Let
`\omega` denote the angular velocity of this satellite. Since this satellite in in a circular motion, the acceleration on this satellite would need to satisfy
`a = \omega^(2) \, r`.

In other words:

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

`\begin{aligned} r &= \left((G\, M)/(\omega^(2))\right)^(1/3)\end{aligned}`.

The question asks for a rotation of
`3* (2\, \pi) = 6\, \pi\; {\text{rad}}` within a day, which is
`24 * 3600\; {\rm s}`. The angular velocity of this satellite should be:

`\begin{aligned}\omega = \frac{6\, \pi}{24 * 3600\; {\rm s}} \end{aligned}`.

Substitute this value into the expression for
`r` and evaluate:

`\begin{aligned} r &= \left((G\, M)/(\omega^(2))\right)^(1/3) \\ &= \left(\frac{(6.67 * 10^(-11)\; {\rm N \cdot m^(2) \cdot kg^(-2)}) * (5.97 * 10^(24)\; {\rm kg})}{((6\, \pi) / (24 * 3600\; {\rm s}))^(2)}\right)^(1/3) \\ &\approx 2.03 * 10^(7)\; {\rm m}\end{aligned}`.

(Note that
`1\; {\rm N} = 1\; {\rm kg \cdot m \cdot s^(-2)}`.)