Question

1) Show the K.E vs. orbital radius R graph for a satellite orbiting the earth.

2) Show the P.E vs. orbital radius R graph for a satellite orbiting the earth.

3) Show the T.E vs. orbital radius R graph for a satellite orbiting the earth.

Answer 1

$1)$ Step 1: Find the orbital velocity of the satellite.
Given:
Mass of earth = M
Radius of orbite of satellite = R
Mass of satellite = m
Let's denote the orbital velocity of the satellite as $v_{\circ }.$
Now as we know, in equilibrium, the gravitational pull provides the necessary centripetal force i.e., force of gravity$\left(F_g\right)$ is equal to the centripetal force$\left(F_c\right).$
Therefore,
Centripetal Force$\left(F_c\right)=\frac{m{v}_0^2}{r}....\left(i\right)$

Force of gravity$\left(F_g\right)=\frac{GMm}{r^2}....\left(ii\right)$

From equation $\left(i\right)\&\left(ii\right)$

$F_g=F_c$

We have,

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

Therefore, orbital velocity of the body $\text{(satellite)}$ is given by,

$v_{\circ }=\sqrt{\frac{GM}{r}}=\sqrt{\frac{GM}{R}}$

Step 2: Find the kinetic Energy of the satellite.
As we know,

$K.E=\frac{1}{2}m{v}_0^2$

By putting the values of $v_{\circ }$ in K.E,

we get,

$K.E=\frac{1}{2}m.\frac{GM}{R}$

$K.E=\frac{GMm}{2R}$

Therefore, as per the equation, kinetic energy is inversely proportional to $R.$

$K.E\propto \frac{1}{R}$

Step 3: Draw $KE$ vs orbital radius $\left(R\right)$ graph.


$2)$ Step 1: Find the relation between potential energy and orbital radius.

As we know,

Gravitational potential energy of the body,

$U=-\frac{GMm}{r}$

$U\propto -\frac{1}{r}$

Therefore, when $r$ invreases, $PE$ becomes less negative i.e., increases.

Step 2: Draw P.E vs. orbital radius R graph.


$3)$ Step 1: Find the equation of total Energy.
As we know,

$T.E.=K.E.+P.E.$

Therefore,

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

$T.E=-\frac{GMm}{2R}$

This shows, $T.E\propto \frac{1}{R}$

Step 2: Draw T.E vs orbital radius R graph.