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. $\dfrac{1}{2}m{v^2}$

B. $\dfrac{1}{4}m{v^2}$

C. $ - \dfrac{1}{4}m{v^2}$

D. $ - m{v^2}$

E. $ - \dfrac{1}{2}m{v^2}$

Answer 1

Hint: Here we have to use the concept of total energy to get the answer.
Total energy is equal to the sum of potential energy and kinetic energy.

Complete step by step answer:
The satellites move around the planet in either a circular way or elliptical way.
The satellites circling about the earth moves in a roundabout movement at a steady speed and at a fixed height by moving with a tangential speed that permits it to fall at a similar rate at which the earth bends. The power of gravity acts toward a path opposite to the course of movement of the satellite all through the direction.
In round movement about the earth, a satellite stays at a fixed distance ways from the outside of the Earth at constant. Since the tangential speed is an element of the radius of the circle, the speed stays consistent along with the kinetic energy. Additionally, since the potential energy is dependent on the height, which stays steady for this situation, along these lines, the potential energy stays consistent all through. Thus, the all out mechanical energy for example KE + PE stays consistent.
The tangential velocity is given by:
$v = \sqrt {\dfrac{{GM}}{{R + h}}} $

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.
So, kinetic energy is given by:
${\text{K}}{\text{.E}} = \dfrac{1}
{2}m{v^2} = \dfrac{{GmM}}
{{2\left( {R + h} \right)}}$

Where $m$ is the mass of the satellite and $v$ is the speed.
The potential energy is given by:
${\text{P}}{\text{.E}} = - \dfrac{{GmM}}
{{\left( {R + h} \right)}}$

Total energy
 $
   = \dfrac{{GmM}}
{{2\left( {R + h} \right)}} - \dfrac{{GmM}}
{{\left( {R + h} \right)}} \\
   \Rightarrow - \dfrac{{GmM}}
{{2\left( {R + h} \right)}} \\
 $

Which is equal to the kinetic negative of kinetic energy.
So, total energy $ = - \dfrac{1}{2}m{v^2}$

Hence, option E is correct.

Note:Here we may get confused between option A and option E but we should remember that the total energy is the sum of both potential and kinetic energy. Since, potential energy is negative, so, the total energy becomes negative.