Why a Satellite Speeds Up When Air Drag Tries to Slow It Down

Throw a ball up and it comes back. Throw it sideways fast enough and it falls around the Earth instead of into it. Faster still, and it never returns.

The exact speeds for these come from gravitational energy, which must be measured from a zero point far away in space.

This part covers gravitational potential energy and potential, escape speed, orbital speed and period, and the energy of a satellite. Take $g=9.8$ m/s${}^2$ and $R=6.4\times 10^6$ m, so $GM=g{R}^2\approx 4.0\times 10^{14}$ m${}^3$/s${}^2$.

**Gravitational potential energy of two masses a distance $r$ apart is $U=-\frac{GMm}{r}$, the work done to bring them together from infinity; gravitational potential $V=-\frac{GM}{r}$ is that energy per unit mass, a property of the point in space alone.

Both are zero at infinity** and negative everywhere else, because gravity attracts. $U=mV$.

Worked example 1 — at the surface. For a $100$ kg satellite on the ground:

$$U=-\frac{GMm}{R}=-mgR=-100\times 9.8\times 6.4\times 10^6\approx -6.27\times 10^9\text{ J}$$

$$V=-gR\approx -6.27\times 10^7\text{ J/kg}$$

**Worked example 2 — lifting to $h=R$.**

$$\Delta U=GMm\left(\frac{1}{R}-\frac{1}{2R}\right)=\frac{mgR}{2}\approx 3.14\times 10^9\text{ J}$$

Using $mgh$ would wrongly give twice this, because $g$ weakens with height.

An everyday example. Carrying a water bucket up to the terrace, $mgh$ works perfectly — the height is tiny compared with Earth's radius.

The substance. Potential belongs to a point; potential energy belongs to a pair of masses.

**A body just escapes when its total energy reaches zero, so $\frac{1}{2}m{v}_e^2-\frac{GMm}{R}=0$, giving $v_e=\sqrt{\frac{2GM}{R}}=\sqrt{2gR}$.

Worked example 1 — Earth.**

$$v_e=\sqrt{2\times 9.8\times 6.4\times 10^6}=\sqrt{1.2544\times 10^8}\approx 1.12\times 10^4\text{ m/s}=11.2\text{ km/s}$$

Worked example 2 — the Moon. With $M=7.4\times 10^{22}$ kg and $R=1.74\times 10^6$ m:

$$v_e=\sqrt{\frac{2\times 6.67\times 10^{-11}\times 7.4\times 10^{22}}{1.74\times 10^6}}\approx 2.4\text{ km/s}$$

Worked example 3 — scaling. A planet with $4$ times Earth's mass and the same radius has $v_e=\sqrt{4}\times 11.2=22.4$ km/s.

An everyday example. A stone thrown upward from a rooftop always falls back, because even the strongest throw is far below $11.2$ km/s.

The substance. Escape speed does not depend on the mass of the body or the direction of launch, ignoring air resistance.

**Gravity supplies the centripetal force, $\frac{m{v}^2}{r}=\frac{GMm}{r^2}$, so the orbital speed is $v_o=\sqrt{\frac{GM}{r}}$ and the period is $T=\frac{2\pi r}{v_o}=2\pi \sqrt{\frac{r^3}{GM}}$, where $r=R+h$.

Worked example 1 — just above the surface** ($r\approx R$):

$$v_o=\sqrt{gR}\approx 7.92\text{ km/s}T=2\pi \sqrt{\frac{R}{g}}\approx 5080\text{ s}\approx 85\text{ min}$$

Note that $v_e=\sqrt{2}{v}_o$ at the same place.

**Worked example 2 — at $600$ km height** ($r=7.0\times 10^6$ m):

$$v_o=\sqrt{\frac{4.01\times 10^{14}}{7.0\times 10^6}}\approx 7.57\text{ km/s}T=\frac{2\pi \times 7.0\times 10^6}{7570}\approx 5810\text{ s}\approx 97\text{ min}$$

Higher orbit, slower speed, longer period — consistent with $T^2\propto r^3$.

An everyday example. A television dish on a rooftop stays pointed at one spot because its satellite orbits once a day above the equator, keeping pace with Earth's rotation.

The substance. Orbital speed does not depend on the satellite's mass — a small and a large satellite at the same height move equally fast.

**A satellite in a circular orbit has $K=\frac{GMm}{2r}$, $U=-\frac{GMm}{r}$ and total energy $E=-\frac{GMm}{2r}$; the total is negative because the satellite is bound to Earth and needs energy added to escape.**

Notice that $K=-E$ and $U=2E$.

Worked example. A $1000$ kg satellite orbits at $r=7.0\times 10^6$ m, with $GM\approx 4.01\times 10^{14}$ m${}^3$/s${}^2$.

$$K=\frac{4.01\times 10^{14}\times 1000}{2\times 7.0\times 10^6}\approx 2.87\times 10^{10}\text{ J}$$

$$U\approx -5.73\times 10^{10}\text{ J}E\approx -2.87\times 10^{10}\text{ J}$$

- Energy to escape from this orbit $=+2.87\times 10^{10}$ J
- **Energy to move to $r=1.4\times 10^7$ m**: $E$ rises to $-1.43\times 10^{10}$ J, so about $1.43\times 10^{10}$ J is needed

An everyday example. A marble sitting in a bowl is bound — it needs a push of energy to climb over the rim, just as a satellite needs energy to leave its orbit.

The substance. Raising a satellite needs energy even though it ends up moving slower — its potential energy rises by twice as much as its kinetic energy falls.