- Practice Questions & MCQ

Orbital velocity of a satellite is the velocity which is required to put the satellite into its orbit around the earth.

For the revolution of a satellite around the earth, there must be a centripetal force for the circular motion to happen.

The gravitational force which acts towards the center is providing the required centripetal force,

i.e $F_c=F_G$

Gravitational force,

$
F_c=\frac{m v^2}{r}
$

Where Centrifugal force,

So on equating we get

$
\begin{aligned}
& \frac{m v^2}{r}=\frac{G M m}{r^2} \\
v= & \sqrt{\frac{G M}{r}}
\end{aligned}
$

Where
$r \rightarrow$ Position of satellite from the centre of earth
$v \rightarrow$ Orbital velocity
- If $r=(R+h)$ where $R$ is the radius of the earth then:

$
v=\sqrt{\frac{g R^2}{R+h}}=R \sqrt{\frac{g}{R+h}} \quad\left[\text { As } G M=g R^2 \text { and } r=R+h\right]
$
 

• Dependence of Orbital velocity

              1. Orbital velocity is independent of the mass of satellite and is always along the tangent of the orbit.

               2.It depends upon the mass of the central body and radius of orbit

                     means, Greater the value of radius of orbit, less be the orbital velocity                    

• If satellite is close to the earth’s surface,

            As $\mathrm{h} \ll \mathrm{R}$ or $h \approx 0$
and using $G M=g R^2$
So $V=\sqrt{\frac{G M}{R}}=\sqrt{g R}$
$V=\sqrt{9.8 \times 6.4 \times 10^6}$
$=7.9 \mathrm{~km} / \mathrm{s} \simeq 8 \mathrm{~km} / \mathrm{s}$
Where
$V \rightarrow$ Orbital velocity
$g \rightarrow 9.8 \mathrm{~m} / \mathrm{s}^2$
$R \rightarrow$ Radius of Earth
- Angular momentum of satellite
$L=m v r$
$L=\sqrt{m^2 G M r}$
$L=$ Angular momentum
$m \rightarrow$ mass of satellite
$v \rightarrow$ Orbital velocity of the satellite

• Relation of escape velocity and orbital velocity

we know that

$
V_e=\sqrt{\frac{2 G M}{R}}
$

$
\begin{aligned}
V & =\sqrt{\frac{G M}{R}} \\
\Rightarrow V & =\frac{V_e}{\sqrt{2}}
\end{aligned}
$

Where
$V \rightarrow$ Orbital velocity
$V_e \rightarrow$ Escape velocity
or

$
V_{\text {escape }}=\sqrt{2} V_{\text {orbital }}
$

Or we can say that
If the speed of satellite is made $\sqrt{2}$ times the original speed, then it will escape from the gravitational pull of the earth.
- Shape of orbit of satellite

If $0<V<v_o$, then satellite does not remain in it's circular path rather it traces a spiral path and falls on earth
$V=v_o \quad$ Satellite revolves in circular path
$V=v_e \quad$ satellite move along the parabolic path and will escape from gravitational pull.
$V>v_e$ satellite will escape but now the of motion will be hyperbolic.

Here, 

$\mathrm{V}=$ velocity of body
$v_o$ - orbital velocity of a body
$v_e-$ escape velocity of a body