Notes on The Motion of Satellite and Black Holes | Grade 11 > Physics > Gravity and Gravitation | KULLABS.COM

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• Note
• Things to remember
• Videos ### The Motion of Satellite

A satellite is a body which revolves continuously around a bigger body. The gravitational force between them provides the required centripetal force. There are two types of satellite and artificial satellite. Moon is the natural satellite of the earth. Artificial satellites are put into their respective orbits with the help of rockets.

#### Orbital Velocity

The velocity required to keep the satellite into its orbit is called orbital velocity. A satellite keeps on revolving round the earth with a certain velocity which depends on the radius of its orbit. This velocity is called the orbital velocity of the satellite.

Suppose a satellite of mass m revolving around the earth at height h form the surface as shown in the figure. Let v0 be the orbital velocity of the satellite. The centripetal force on the planet is

Fc = $$\frac{mv^2}{r}$$=$$\frac{mv^2}{R+H}$$

And the gravitational force between the earth and the satellite is

Fa= $$\frac{GMm}{r^2}$$ =$$\frac{GMm}{(R+h)^2}$$ where M is the mass of the earth .As the gravitational force provides the required centripetal force for the satellite in its orbit then

Fa = Fc

$$\frac{GMm}{r^2}$$ =$$\frac{mv^2}{r}$$

v0= $$\sqrt{\frac{GM}{R}}$$

= $$\sqrt{\frac{GM}{R+h}}$$ ......................(i)

if g is the acceleration due t gravity on the earths surface then ,

g = $$\frac{GM}{R^2}$$

GM = gR2

and V0 = $$\sqrt\frac{gR^2}{r}$$ =R $$\sqrt\frac{g}{r}$$

This is the expression for orbital velocity of the satellite .

if the satellite is moving very close to earth's surface ,we can neglect h and

v0 = $$\sqrt{gR}$$

V0= $$\sqrt{9.81×6400×1000}$$

= 7.2×103 ms-1

= 7.2kms-1

Period of Satellite

The period, T of a satellite is the time required to complete one revolution round the earth.

∴ T =$$\frac{circumstance of circular orbit}{orbit velocity}$$

o T = $$\frac{2πr}{v_{0}}$$ =2πr$$\sqrt\frac{r}{GM}$$ =2π(R+h)$$\sqrt\frac{R+h}{GM}$$

If the satellite is very close to the earth’s surface, then

\begin{align*} T &= 2\pi \sqrt {\frac {R^3} {GM} } \\ \text {or,} \: T &= 2\pi \sqrt {\frac Rg} \\ R &= 6.4 \times 10^6 m \\ g&= 9.8 ms^{-1} \\ \therefore T &= 2\pi \sqrt {\frac {6.4\times 10^6}{9.8} } = 5000 s = 84 \text {minutes approximately} \\ \end{align*}

Height of Satellite (h)

It is the distance between earth and satellite. We know

\begin{align*} T &= 2\pi \sqrt { \frac {(R+h)^2}{GM}} \\ \text {Squaring both sides, we get} \\ T^2 &= 4\pi ^2 \frac {(R+h)^3}{GM} \\ \text {or,} \: (R+h)^3 &= \frac {T^2GM}{4\pi ^2} \\ \text {Taking cube root in both sides} \\ (R+h) &= \left (\frac {T^2GM}{4\pi ^2} \right )^{1/3}\\ \therefore h &= \left (\frac {T^2GM}{4\pi ^2} \right )^{1/3} – R \\ \end{align*}

#### Geostationary Satellite

The satellite which appears stationary above from the earth’s surface is called geostationary satellite. It is because the time period of the satellite around the earth is same as the time taken by the earth to complete one rotation on its axis. The orbit of the geostationary satellite is called

parking orbit. Height, h of the geostationary satellite can be calculated as follows.

T=$$\frac{2π}{R}$$$$\sqrt\frac{(R+h)^2}{g}$$

on simplyfying,

h={$$\frac{T^2R^2×g}{4π^2}$$}1/3 -R

substituting the values R= 6.4×106m

T=24×3600 sec

g= 9.8 ms-2

in equn (iv) we get

h= {$$\frac{(24×3600)^2×(6.4×10^6)^2×9.8}{4π^2}$$}1/3-6.4×106 =3.6×107=36,000 km

radius of parkiung orbit ,r=R+h =6.4×103 km+36000 km = 42,4000 km

Speed of satellite

Thre speed of satellite at the parking orbit is given by ,

v0=$$\frac{2πr}{T}$$

T=24×3600 s and r=36000+6400 n= 42,400 km

∴ v0= $$\frac{2π×42400}{24×3600}$$ = 3.,1×103 m/s =3.1 km/s

#### Energy of a Satellite

A satellite revolving round the earth has both the kinetic energy and potential energy. Suppose a satellite of mass m is at a height of h from the surface of mass M and radius R. The radius R of the orbit of the satellite is R+h.

The gravitation force between the earth and the satellite is

Fg= $$\frac{GMm}{(R+h)^2}$$

which provide the required centripetal force to the satellite .if v0 is the orbital velocity of the satellite,then the centripetal force is Fc=$$\frac{mv_{0}}{(R+h)}$$

= $$\frac{GMm}{(R+h)^2}$$ = $$\frac{mv_{0}}{(R+h)}$$

or V02=$$\frac{GM}{R+h}$$

∴ Kinetic energy = $$\frac{1}{2}$$mv02=$$\frac{1}{2}$$m$$\frac{GM}{(R+h)}$$

or K.E.= $$\frac{GMm}{2(R+h)}$$

Total potential energy of the satellite at the height h above the earths surface is

P.E. =- $$\frac{GMm}{(R+h)}$$

∴ Total energy = P.E. + k.E.

E =$$\frac{GMm}{2(R+h)}$$ +- $$\frac{GMm}{(R+h)}$$

E =- $$\frac{GMm}{2(R+h)}$$

The negative sign in above equation means that the satellite is bounded to the earth .i.e. energy has to supply from outside to free a satellite from its orbit .

### Black Holes

If a star of mass more than three solar masses has completely burned its nuclear fuel, it should collapse into configuration known as black hole. The resulting object is independent of the properties of matter that produces it and is completely described by its mass and completely encloses the collapsed matter. The horizon is an ideal one way membrane, i.e. particles and light can go inward through the surface but cannot come outward. As a result, the object is dark i.e. black and hides from view a finite region of space.

The escape velocity, $$v = \sqrt {\frac {2GM}{R} }$$ shows that the body of mass will acts as a black hole if its radius R is less than or equal to a certain critical radius. Karl Schwarzschild in 1916, derived an expression for the critical velocity from Einstein’s general theory of relativity, known as Schwarzschild radius (Rs). Schwarzschild radius is given below

\begin{align*} c &= \sqrt {\frac {2GM}{R_s}} \\ \text {or,} R_s &= \frac {2GM}{c^2} \\ \end{align*}

If a spherical, non-rotating body of mass M has a radius smaller than Rs, then nothing even light cannot escape from the surface of the body. The body is then black hole. Any other body within a distance Rs from the centre of the black hole is trapped by the gravitational attraction of the black hole and cannot escape from it.

The centripetal force on the planet is

Fc = $$\frac{mv^2}{r}$$=$$\frac{mv^2}{R+H}$$

Thre speed of satellite at the parking orbit is given by ,

v0=$$\frac{2πr}{T}$$

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##### Ranu Dhungel

characteristic feature of black hole