Question

Magnitude of binding energy of satellite is E and kinetic energy is K. The ratio $\dfrac{E}{K}$ is:
$\text{A}\text{. }\dfrac{2}{1}$
$\text{B}\text{. }\dfrac{1}{4}$
$\text{C}\text{. 1}$
$\text{D}\text{. }\dfrac{1}{2}$

Answer 1

Hint: Binding energy (E) is the sum of the potential energy and the kinetic energy of the satellite. The potential energy of the satellite is $U=-\dfrac{GmM}{r}$. The kinetic energy of the satellite will be $K=\dfrac{1}{2}m{{v}^{2}}$ and v =$\sqrt{\dfrac{GM}{r}}$. Use these formulas to find the expression for the binding energy. Then divide E by K to find the ratio.

Complete step by step answer:
A satellite revolves around the earth under the influence of gravity. The earth exerts a force called gravitational force on the satellite.
If the mass of the satellite is m and it is revolving in an orbit of radius r then it is attracted towards the earth by a force equal to $F=G\dfrac{mM}{{{r}^{2}}}$.
Here, M is the mass of the earth and G is the universal gravitational constant.
We say that the satellite is bound to the earth.
When the satellite is revolving around the earth, it possesses some kinetic energy and some potential energy.
Suppose a satellite of mass m is revolving around the earth with a speed v, in an orbit of radius r and mass of the earth is M.
The potential energy of this satellite is given as $U=-\dfrac{GmM}{r}$.
We know that kinetic energy of the satellite will be $K=\dfrac{1}{2}m{{v}^{2}}$ ….. (i).
For a revolving body under gravitational force, the value of speed v is equal to $\sqrt{\dfrac{GM}{r}}$.
Substitute the value of speed v in equation (i).
$K=\dfrac{1}{2}m{{\left( \sqrt{\dfrac{GM}{r}} \right)}^{2}}$
$\Rightarrow K=\dfrac{1}{2}m\dfrac{GM}{r}=\dfrac{1}{2}\dfrac{GmM}{r}$.
The sum of the kinetic energy and potential energy of the satellite is its total energy called the binding energy of the satellite.
Therefore, the binding energy (B.E) of the satellite will be equal to K+U.
This implies, $E.B=K+U=\dfrac{1}{2}\dfrac{GmM}{r}-\dfrac{GmM}{r}=-\dfrac{1}{2}\dfrac{GmM}{r}$ .
The magnitude of binding energy will be $\dfrac{1}{2}\dfrac{GmM}{r}$ and this value is equal to E.
As you can see, E and K are equal.
Therefore, the ratio $\dfrac{E}{K}$ will be 1.
Hence, the correct option is C.

Note: The negative sign in the formula of potential energy denotes that the nature of force is attractive, which means that the earth is attracting the satellite towards itself.
If the force is repulsive then, the potential energy will be a positive value.