MOTION OF A SATELLITE IN A CIRCULAR

Orbital Speed

Suppose that a satellite of mass m orbits a planet in a circular orbit of radius r.

$\Rightarrow$ F_g= $\frac{GMm}{r^2}$

If v_o is the orbital velocity of the satellite, then,

$\frac{m{v}_0^2}{r}$= $\frac{GMm}{r^2}$

$\Rightarrow$ v₀= $\sqrt{\frac{GM}{r}}$ …(1)

Putting $\frac{GM}{R^2}$= g (acceleration due to gravity on the surface), we obtain,

$\Rightarrow$ v₀= $\sqrt{\frac{g{R}^2}{r}}$ …(2)

If the altitude of the orbit is h, then,

$\Rightarrow$ v₀= $\sqrt{\frac{GM}{R+h}}$ = $\sqrt{\frac{GM}{R(1+h/R)}}$ = $\sqrt{\frac{gR}{(1+h/R)}}$.

Angular Speed

The corresponding angular speed is,

$\omega$ = v₀/r

Putting v₀ = $\sqrt{\frac{GM}{r}}$, we obtain,

$\omega$ = $\sqrt{\frac{GM}{r^3}}$ = $\sqrt{\frac{GM}{{(R+h)}^3}}$.

Time Period of Revolution

The period of revolution

T = 2$\pi$/$\omega$

Putting, $\omega$ = $\sqrt{\frac{GM}{r^3}}$,

T = 2$\pi$ $\sqrt{\frac{r^3}{GM}}$ = 2$\pi$ $\sqrt{\frac{{(R+h)}^3}{GM}}$.

Potential Energy

The Gravitational potential energy of the planet satellite system is

PE = -$\frac{GMm}{r}$.

The -ve sign indicates that the satellite is subject to planet's gravitational field.

Kinetic Energy

KE of a satellite in orbit is given by

KE = $\frac{1}{2}$ $m{v}_0^2$, where v₀ = $\sqrt{\frac{GM}{r}}$,

KE = $\frac{GMm}{2r}$.

Total Energy

The total mechanical energy of the satellite is equal to the sum of its potential and kinetic energies.

Total energy E = $\left(-\frac{GMm}{r}\right)$ + $\left(\frac{GMm}{2r}\right)$ = -$\frac{GMm}{2r}$

The total energy is negative because the satellite is bound by the planet's gravitational field.

Angular Momentum

The angular momentum of the satellite is given by

L = mv₀r.

Putting v₀ = $\sqrt{\frac{GM}{r}}$, we obtain, L = m($\sqrt{GMr}$).

ESCAPE VELOCITY

It is the minimum velocity required to escape from the gravitational field of earth.

When a rocket is fired with the minimum escape speed $v_{esc}$ it will reach $r=\infty$ with zero speed; that is, $E_f=K+U=0.$ Its initial energy at the earth's surface is

$E_i=\frac{1}{2}m{v}_{esc}^2-\frac{GMm}{R}$

Using energy conservation

$E_f=E_i$

$0=\frac{1}{2}m{v}_{esc}^2-\frac{GMm}{R}$

or $v_{esc}^{}=\sqrt{\frac{2GM}{R}}$

or $v_{esc}=\sqrt{2gR}$

Illustration 1:

A body is projected vertically upwards from the surface of earth with a velocity sufficient to carry it to infinity. Calculate the time taken by it to reach height h.

Solution: If at a distance r from the centre of the earth the body has velocity v, by conservation of mechanical energy,

$\frac{1}{2}m{v}^2+\left(-\frac{GMm}{r}\right)=\frac{1}{2}m{v}_e^2+\left(-\frac{GMm}{R}\right)$

or $v^2=v_e^2+\frac{2GM}{R}\left[\frac{R}{r}-1\right]$

But as $v_e=\sqrt{2gR}$ and $g=\left(GM/R^2\right)$

$v^2=2gR+2gR\left[\left(R/r\right)-1\right]$

or $v=\sqrt{\frac{2g{R}^2}{r}},$ i.e., $\frac{dr}{dt}=R\frac{\sqrt{2g}}{\sqrt{r}}$

i.e. $\underset{0}{\overset{t}{\int }}dt=\frac{1}{R\sqrt{2g}}\underset{R}{\overset{R+h}{\int }}{r}^{1/2}dr$

i.e., $t=\frac{2}{3}\frac{1}{R\sqrt{2g}}\left[{\left(R+h\right)}^{3/2}-R^{3/2}\right]$

i.e., $t=\frac{1}{3}\sqrt{\frac{2R}{g}}\left[{\left(1+\frac{h}{R}\right)}^{3/2}-1\right]$

BOUND AND UNBOUND TRAJECTORIES

If v is the velocity given to a satellite. And V_c is the velocity for a circular orbit and V_e is the escape velocity.

$V_c=\sqrt{\frac{GM}{r}}$ and $Ve=\sqrt{\frac{2GM}{r}}$

Where r is the distance of the satellite from centre of the earth.

1. v < V_c - The satellite follows an elliptical path with centre of earth as the farther focus. In this case, if satellite is projected from near the surface of earth, it will hit the earth's surface without completing the orbit.

2. v = V_c - The satellite follows a circular orbit with the centre of earth as the centre of orbit.

3. V_c < v < V_e - The satellite follows an elliptical orbit with the centre of earth as the focus nearer to the point of projection.

4. v = V_{e -} The satellite escapes from the field of earth along a parabolic trajectory.

5. v > V_e - The satellite escapes the field of earth along a hyperbolic trajectory.

GEO STATIONARY SATELLITES

(i) The time period of the satellite around the earth must be equal to the rotational period of the earth (i.e. 24 hours).

(ii) The direction of motion of the satellite must be same as that of the earth.

Illustration 2:

A body is orbiting around the earth at mean radius 9 times as great as the orbit of a geostationary satellite. In how many days it will complete one revolution around the earth and what is its angular velocity?

Solution: $(T_1^2/T_2^2)=(R_1^3/R_2^3)$

$\Rightarrow$ $T_2^2={\left(\frac{R_2}{R_1}\right)}^3{T}_1^2$ $\Rightarrow$ $T_2={\left(\frac{R_2}{R_1}\right)}^{3/2}{T}_1$

For the geostationary satellite, T₁ = 1 day (24 hrs.)

$\therefore$ T₂ = (9R³ / R³)^½ = 27days

So it will complete one revolution in 27 days

$\omega$ = 2$\pi$/T₂ = $\frac{2\pi }{27\times 24\times 3600}$rad/s = 2.693 ´ 10^-6 rad/sec.

KEPLER'S LAWS

1. First law (Law of orbit) : Each planet moves in an elliptical orbit, with the sun at one focus of the ellipse. This is called law of orbits.

2. Second law (Law of area): A line joining any planet to the sun (i.e. radius vector of the planet form sun) sweeps equal areas in equal interval of time

Third law (Law of period) : The square of the time periods of the planets are proportional to the cube of semi-major axis of ellipse.

Illustration 3:

A planet of mass m moves along an ellipse around the sun so that its maximum and minimum distances from the sun are equal to R and r respectively. Find the angular momentum of this planet relative to the centre of the sun.

Solution: According to Kepler's Second Law the angular momentum of the planet is constant, we have

mv₁R = mv₂r, v₁R = v₂r

If the mass of the Sun is M conserving total mechanical energy of the system at two given positions we have,

– $\frac{GMm}{R}+\frac{1}{2}m{v}_1^2=-\frac{GMm}{r}+\frac{1}{2}m{v}_2^2$.

$\therefore GM\left[\frac{1}{R}-\frac{1}{r}\right]=\frac{v_2^2}{2}-\frac{v_1^2}{2}$

Or $GM\left[\frac{r-R}{Rr}\right]=\frac{v_1^2{R}^2}{2r^2}-\frac{v_1^2}{2}$

Or $GM\left[\frac{r-R}{Rr}\right]=\frac{v_1^2}{2}\left(\frac{R_{}^2}{r^2}-1\right)$$=\frac{v_1^2}{2}\left(\frac{R^2-r^2}{r^2}\right)$

$\therefore v_1^2=\frac{2GM\left(R-r\right)r^2}{Rr\left(R^2-r^2\right)}=\frac{2GMr}{R\left(R+r\right)}$

Or $\sqrt{\frac{2GMr}{R\left(R+r\right)}}$

Now angular momentum = mv₁R = m$\sqrt{\frac{2GMr}{R\left(R+r\right)}}$

AT A GLANCE

1. Newton's Law $F=\frac{G{m}_1{m}_2}{r^2}$; $6.67\times 10^{-11}N{m}^2/k{g}^2$; where $m_1$ and $m_2$ are point masses.

2. Variation of acceleration due to gravity

(a) inside the earth $g=g_0\frac{r}{R}$

(b) outside the earth $g=g_0{\left(\frac{r}{R}\right)}^2$

Where $g_0$ is the acceleration due to gravity on the surface of earth

$g_0=\frac{GM}{R^2}$, R = radius of earth.

3. Variation of acceleration due to gravity close to the earth's surface

$g=g_0=\left[1-\frac{h}{R}\right]$ (inside the earth)

$g=g_0=\left[1-\frac{2h}{R}\right]$ (outside the earth)

h is the small distance measured from the earth's surface.

4. Variation of gravity with latitude a. (due to rotation of earth)

$g^{\prime }=g-{\omega }^{2}Rco{s}^2\alpha$

$\omega =\frac{2\pi }{T}$; T = 24 hours.

$g^{\prime }=g-0.034co{s}^2\alpha$

Where g is the gravity at poles.

5. Gravitational energy of a system of point masses

$u=-\frac{1}{2}\sum _{i\ne j}G\frac{m_i{m}_j}{r_{ij}}$

Eg. For a system of four identical point masses located at corners of a square of side length 'a'.

$U=-\frac{GMm}{r},KE=\frac{1}{2}\frac{GMm}{r}$

6. For a satellite of mass 'm' moving in a circular orbit of radius 'r' about a central body of mass M.

Potential energy $=-\frac{GMm}{r},KE=\frac{1}{2}\frac{GMm}{r}$

Total energy $=-\frac{1}{2}\frac{GMm}{r}$

7. Escape velocity of a body is the velocity at which total energy of the body becomes zero. at a distance 'r' (>R_e) from centre of earth

$V_{escape}=\sqrt{\frac{2G{M}_e}{r}}$