Motion of artificial satellite around the Earth

ORBITAL VELCOITY:

“The velocity required to keep the satellite into its orbit is called ‘Orbital Velocity’ ”.

DERIVATION:

The gravitational pull of Earth on a satellite provides the necessary centripetal force for orbital motion. Since this force is equal to the weight of satellite, ‘$W_S= mg$’, therefore:

$F_C=W_S$ …….. (i)

Since,

$W_S =mg_h$

Put in equation (i):

$\boxed{F_C =mg_h}$ ………. (ii)

Here,

$m=$ Mass of satellite.

$g_h=$ Acceleration due to gravity at height ‘$h$’ from the surface of Earth.

We Know that:

$\boxed{F_C=\frac{mV^2}{r}}$

Compare it with equation (ii), we get:

$mg_h=\frac{mV^2}{r}$

$V^2=g_h \thinspace r$

Take square root on both sides.

$V=\sqrt{g_h \thinspace r}$

Since,

$r= R+h$

Therefore,

$\boxed{V=\sqrt{g_h \thinspace (R+h)}}$ ……….. (iii)

IF h<< R:

If satellite is orbiting very close to the surface of Earth then: h << R In this case orbital radius may be considered equal to radius of Earth. Therefore,

$R + h = R$ Also,
$g = g_h$ & $V = V_C$

For this case equation (iii) becomes:

$\boxed{V_C=\sqrt{gR}}$

Where,

$V_C=$ Critical Velocity.

$g=$ Acceleration due to gravity on the surface of Earth.

CRITICAL VELOCITY:

It can be defined as follows:

“ The constant horizontal velocity required to put the satellite into a stable circular orbit around the Earth is called ‘Critical velocity’ “.

• It is also known as orbital speed or proper speed

VALUE OF $V_C$:

$g= 10 \thinspace m/s^2$

$R= 6.38 \thinspace \text{x} \thinspace 10^6 \thinspace m$

Therefore,

$V_C=\sqrt{(10)( 6.38 \thinspace \text{x} \thinspace 10^6 \thinspace )}$

$V_C =7.99 \thinspace \text{x} \thinspace 10^3 \thinspace m/s$

$\boxed{V_C=8 \thinspace kms^{-1}}$