Why a Geostationary Satellite Seems to Hang Still Over India

The same force that pulls a mango from a tree keeps the Moon circling the Earth and India's satellites in orbit.

This lesson covers Kepler's laws and the law of gravitation, the variation of g, escape and orbital velocity, and geostationary and polar satellites.

**Kepler's laws state that planets move in ellipses with the Sun at one focus, sweep out equal areas in equal times, and obey $T^2\propto a^3$, while Newton's law states that any two masses attract with a force $F=\frac{G{m}_1{m}_2}{r^2}$.

Kepler's laws:

- Law of orbits — each planet moves in an ellipse with the Sun at one focus
- Law of areas — the line from the Sun to a planet sweeps equal areas in equal times, so a planet moves fastest when nearest the Sun
- Law of periods** — $T^2\propto a^3$, where a is the semi-major axis

Newton's law of gravitation:

$$F=\frac{G{m}_1{m}_2}{r^2}G=6.67\times 10^{-11}{\text{ N m}}^2{\text{ kg}}^{-2}$$

The force is always attractive, acts along the line joining the masses, and gives $g=\frac{GM}{R^2}$ at the Earth's surface.

Worked example — Kepler's third law. Mars orbits about 1.52 times as far from the Sun as the Earth, so one Mars orbit takes $1.52^{3/2}=1.87$ Earth orbits.

An everyday example. The sea at Mumbai rises and falls twice a day mainly because of the Moon's gravitational pull on the oceans.

The substance. Gravity between everyday objects is tiny — it becomes noticeable only when one of the masses is enormous, like the Earth.

**The acceleration due to gravity falls with height as $g_h\approx g\left(1-\frac{2h}{R}\right)$ for small heights, and with depth as $g_d=g\left(1-\frac{d}{R}\right)$, reaching zero at the Earth's centre.

Altitude.** At height h above the surface:

$$g_h=\frac{GM}{(R+h{)}^2}=g{\left(1+\frac{h}{R}\right)}^{-2}\approx g\left(1-\frac{2h}{R}\right)\text{for }h\ll R$$

Depth. At depth d, only the inner sphere of radius $R-d$ attracts the body. For uniform density $\rho$, with $g=\frac{4}{3}\pi G\rho R$:

$$g_d=\frac{G\times \frac{4}{3}\pi (R-d{)}^3\rho }{(R-d{)}^2}=\frac{4}{3}\pi G\rho (R-d)=g\left(1-\frac{d}{R}\right)$$

Worked example. With $g=9.8$ m s${}^{-2}$ and $R=6400$ km:

- At a height of 32 km: $g_h=9.8\left(1-\frac{64}{6400}\right)=9.70$ m s${}^{-2}$
- At a depth of 64 km: $g_d=9.8\left(1-\frac{64}{6400}\right)=9.70$ m s${}^{-2}$

The same fall in g needs twice the distance below the surface as above it.

An everyday example. A packet of rice on a spring balance at a high Himalayan pass shows a slightly smaller weight than in the plains, though its mass is unchanged.

The substance. g is greatest at the Earth's surface — it falls as $1/r^2$ above the surface and in direct proportion to distance from the centre below it.

**Gravitational potential energy at distance r from the Earth's centre is $U=-\frac{GMm}{r}$; escape velocity is $v_e=\sqrt{2gR}$, and a satellite orbiting close to the Earth needs $v_o=\sqrt{gR}$.

Potential energy.** $U=-\frac{GMm}{r}$ is zero at infinity and negative elsewhere, because gravity attracts; near the surface, the change $mgh$ is a good approximation.

Escape velocity — the smallest launch speed to leave the Earth's pull:

$$\frac{1}{2}m{v}_e^2=\frac{GMm}{R}\Rightarrow v_e=\sqrt{\frac{2GM}{R}}=\sqrt{2gR}$$

Orbital velocity — gravity supplies the centripetal force:

$$\frac{m{v}_o^2}{r}=\frac{GMm}{r^2}\Rightarrow v_o=\sqrt{\frac{GM}{r}},\text{ which is }\sqrt{gR}\text{ near the surface}$$

Worked example. With $g=9.8$ m s${}^{-2}$ and $R=6.4\times 10^6$ m:

$$v_e=\sqrt{2\times 9.8\times 6.4\times 10^6}=11.2{\text{ km s}}^{-1}{v}_o=\sqrt{9.8\times 6.4\times 10^6}=7.9{\text{ km s}}^{-1}$$

An everyday example. Rockets carrying Indian spacecraft towards the Moon must build up speeds close to escape velocity to break free of the Earth's pull.

The substance. Escape velocity does not depend on the mass launched or the direction of launch — a pebble and a spacecraft need the same 11.2 km s${}^{-1}$, ignoring air.

A geostationary satellite orbits above the equator at about 36 000 km, once every 24 hours in the direction the Earth rotates, so it appears fixed in the sky, while a polar satellite orbits much lower over the poles and scans the whole Earth strip by strip.

Geostationary satellites:

- Stay above one point on the equator, so ground dishes need no tracking
- Used for television, telecommunication and weather monitoring, as by India's INSAT and GSAT satellites

Worked example — the height of the orbit. From $T^2=\frac{4{\pi }^2{r}^3}{GM}$, with $T=86400$ s and $GM=g{R}^2=9.8\times (6.4\times 10^6{)}^2$:

$$r={\left(\frac{GM{T}^2}{4{\pi }^2}\right)}^{1/3}=4.2\times 10^7\text{ m}$$

Subtracting the Earth's radius gives a height of about $3.6\times 10^7$ m, or 36 000 km.

Polar satellites:

- Orbit at a few hundred kilometres, passing over both poles
- The Earth turns beneath the orbit, so successive passes cover the whole surface
- Used for remote sensing, mapping, crop and forest monitoring, and weather studies

An everyday example. A DTH dish on a rooftop in any Indian town points in one fixed direction because the geostationary satellite it receives from never moves across the sky.

The substance. A geostationary satellite must be above the equator — a 24-hour orbit tilted to the equator would drift north and south in the sky each day.