Question

A satellite orbits a planet of unknown mass in a circular orbit of radius 2.3 x 104 km. The gravitational force on the satellite from the planet is 6600 N. What is the kinetic energy of the satellite

Answer 1

Answer:

The  kinetic energy is $KE = 7.59 *10^{10} \ J$

Explanation:

From the question we are told that

       The  radius of the orbit is  $r = 2.3 *10^{4} \ km = 2.3 *10^{7} \ m$

       The gravitational force is  $F_g = 6600 \ N$

The kinetic energy of the satellite is mathematically represented as

       $KE = \frac{1}{2} * mv^2$

where v is the speed of the satellite which is mathematically represented as

     $v = \sqrt{\frac{G M}{r^2} }$

=>  $v^2 = \frac{GM }{r}$

substituting this into the equation

      $KE = \frac{ 1}{2} *\frac{GMm}{r}$

Now the gravitational force of the planet is mathematically represented as

      $F_g = \frac{GMm}{r^2}$

Where M is the mass of the planet and  m is the mass of the satellite

 Now looking at the formula for KE we see that we can represent it as

     $KE = \frac{ 1}{2} *[\frac{GMm}{r^2}] * r$

=>    $KE = \frac{ 1}{2} *F_g * r$

substituting values

       $KE = \frac{ 1}{2} *6600 * 2.3*10^{7}$

         $KE = 7.59 *10^{10} \ J$