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

Answer:

Approximately $2.03 \times 10^{7}\; {\rm m}$.

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:

$\begin{aligned} \frac{G\, M}{r^{2}} = a = \omega^{2} \, r \end{aligned}$.

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

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

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

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

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

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