Problem 71
Hard Gravitational Field

(a) What velocity must an artificial Earth satellite have to…

(a) What velocity must an artificial Earth satellite have to move in a circular orbit at a height $H$ above the Earth's surface? (b) What is its period? (c) What is the first cosmic velocity (the orbital velocity at the Earth's surface), and what is the corresponding period?

Explanation of Method

Circular orbit speed and first cosmic velocity.

Step 1: Idea

A satellite is a body in permanent free fall. It falls toward the Earth all the time, but moves sideways fast enough that the curved surface drops away beneath it just as fast. Mathematically: gravity supplies the centripetal force for a circle of radius $R + H$.

Step 2: (a) Velocity

\[ m\frac{v^2}{R + H} = \gamma\frac{Mm}{(R + H)^2}. \]

The satellite's mass cancels. Using $\gamma M = gR^2$, with $g$ the surface value: $v = \sqrt{\frac{\gamma M}{R + H}} = R\sqrt{\frac{g}{R + H}}$.

The higher the orbit, the slower the satellite.

Step 3: (b) Period

The period is the length of the orbit divided by the speed: $T = \frac{2\pi(R + H)}{v}$.

\[T = 2\pi\sqrt{\frac{(R + H)^3}{\gamma M}}\]

This is Kepler's third law: $T^2$ is proportional to the cube of the orbit radius.

Step 4: (c) At the surface ($H = 0$)

The first cosmic velocity is the speed of a (hypothetical) satellite skimming the surface, ignoring air and mountains: $v_1 = \sqrt{gR} = \sqrt{9.81 \cdot 6.37 \cdot 10^6}$.

\[ T_1 = 2\pi\sqrt{\frac{R}{g}}. \]

\[\begin{gathered} v_1 \approx 7.9\ \mathrm{km/s} \\ T_1 \approx 5060\ \mathrm{s} \approx 84\ \mathrm{min} \end{gathered}\]

This is the smallest speed with which any body can orbit the Earth. Real low satellites (e.g. at 400 km) move at about 7.7 km/s with periods of about 92 min.

Check

$ 2\pi R/v_1 $ $= 2\pi \cdot 6.37 \cdot 10^6/7905 $ $= 5063\ \mathrm{s}\ \checkmark $

Units: $ \sqrt{\mathrm{m/s^2} \cdot \mathrm{m}} $ $= \mathrm{m/s}\ \checkmark $

$v$ $H$ $R$ $v = \sqrt{\frac{\gamma M}{R + H}}$ $T = 2\pi\sqrt{\frac{(R + H)^3}{\gamma M}}$ $v_1 = 7.91\ \mathrm{km/s},\ T_1 \approx 84\ \mathrm{min}$
Satellite on a circular orbit at height $H$ (height exaggerated). Gravity supplies the centripetal force; for $H \to 0$ the speed becomes the first cosmic velocity.

Result

\[\boxed{\,v = \sqrt{\frac{\gamma M}{R + H}} = R\sqrt{\frac{g}{R + H}}\,}\]

\[\boxed{\,T = 2\pi\sqrt{\frac{(R + H)^3}{\gamma M}}\,}\]

\[\boxed{\,\begin{gathered} v_1 \approx 7.9\ \mathrm{km/s} \\ T_1 \approx 5060\ \mathrm{s} \approx 84\ \mathrm{min} \end{gathered}\,}\]