Question

The total energy of an artificial satellite of mass $m$ revolving in a circular orbit around the earth with a speed $v$ is

• A. ${\displaystyle \frac{1}{2}}m{v}^2$
• B. ${\displaystyle \frac{1}{4}}m{v}^2$
• C. $-{\displaystyle \frac{1}{4}}m{v}^2$
• D. $-m{v}^2$
• E. $-{\displaystyle \frac{1}{2}}m{v}^2$

Answer 1

The correct option is E

$-{\displaystyle \frac{1}{2}}m{v}^2$

Explanation for the correct answer:-

Step 1. Given Data,

Mass of artificial satellite=$m$

An artificial satellite revolving in a circular orbit around the earth with a speed is $v$

Step 2. Formula used,

The satellites revolve around the planet in either a circular way or an elliptical way.

The centripetal force of acting on the satellite is balanced by the gravitational force due to earth,

$$\begin{gathered} F_{centripetal}=F_{gravitational} \\ \Rightarrow \frac{m{v}^2}{R+h}=\frac{GM}{{(R+h)}^2} \end{gathered}$$
The tangential velocity is given by:
$v=\sqrt{{\displaystyle \frac{GM}{R+h}}}$

Kinetic energy, $\text{K}\text{.E}=\frac{1}{2}m{v}^2$

Potential energy, $\text{P}\text{.E}=-{\displaystyle \frac{GmM}{\left(R+h\right)}}$

$\text{P}\text{.E}=-m{v}^2$

Total Energy=$PE+KE$

$v$ is the speed.

$m$ is the mass of artificial satellite
Where $G$is the gravitational constant, $M$ is the mass of the earth, $R$ is the radius of the earth and $h$ is the height from the surface of the earth.

Step 3: Calculating the kinetic energy,
$\text{K}\text{.E}={\displaystyle \frac{1}{2}}m{v}^2={\displaystyle \frac{GmM}{2\left(R+h\right)}}$
Where, $m$ is the mass of the satellite and $v$ is the speed.

Step 4: Calculating the potential energy,
$\text{P}\text{.E}=-{\displaystyle \frac{GmM}{\left(R+h\right)}}$

Where $M$ is the mass of the earth, $R$ is the radius of the earth, $h$ is the height from the surface of the earth, $m$ is the mass of the satellite.

Step 5: Calculating the total energy of an artificial satellite,
Total energy,

$$\begin{gathered} ={\displaystyle \frac{GmM}{2\left(R+h\right)}}-{\displaystyle \frac{GmM}{\left(R+h\right)}} \\ \Rightarrow -{\displaystyle \frac{GmM}{2\left(R+h\right)}} \end{gathered}$$
Which is equal to the kinetic negative of kinetic energy. We know $v=\sqrt{{\displaystyle \frac{GM}{R+h}}}$, therefore total energy$=-{\displaystyle \frac{1}{2}}m{v}^2$

So the total energy of an artificial satellite of mass $m$ revolving in a circular orbit around the earth with a speed $v$ is$-\frac{1}{2}m{v}^2$
Hence, option E is the correct answer.