A 50 kg satellite circles the Earth in an orbit with a period of 120 min. What minimum energy is...


A {eq}50 \ kg {/eq} satellite circles the Earth in an orbit with a period of {eq}120 \ min {/eq}.

What minimum energy is required to change the orbit to another circular orbit with a period of {eq}180 \ min {/eq}?

(Earth: radius = {eq}6.4*10^6 \ m {/eq}, mass ={eq}6.0*10^{24} \ kg) {/eq}

The Energy of a Satellite:

{eq}\\ {/eq}

A satellite possesses energy in the form of kinetic energy due to its motion, as well as gravitational potential energy due to the gravitational field of the orbiting planet. The sum of the kinetic energy and the potential energy of a satellite is equal to the total energy of the satellite.

In the case of circular orbits, the kinetic energy, as well as the potential energy, (and therefore, the total energy) depends on the following parameters:

  • The mass of the satellite.
  • The mass of the planet orbited by the satellite.
  • The distance of the satellite from the center of the spherical planet.

The period of a satellite, i.e the time taken to complete one orbit, of a satellite around a spherically shaped planet depends on all of the above parameters except the mass of the satellite.

Answer and Explanation:

{eq}\\ {/eq}

We are given:

  • The mass of the satellite, {eq}m=50\;\rm kg {/eq}
  • The initial period of the satellite, {eq}T_1=120\;\rm min=7.2\times 10^{3}\;\rm s {/eq}
  • The final period of the satellite, {eq}T_2=180\;\rm min=1.08\times 10^{4}\;\rm s {/eq}
  • The mass of the Earth, {eq}M=6\times 10^{24}\;\rm kg {/eq}
  • The radius of the Earth, {eq}R=6.4\times 10^{6}\;\rm m {/eq}

The time period of the motion of a satellite around a planet depends only on the mass, {eq}M {/eq}, of the planet and its distance, {eq}r {/eq}, from the center of the planet by the following equation:

{eq}T=2\pi\sqrt {\dfrac{r^3}{GM}} {/eq}


  • {eq}G=6.67\times 10^{-11}\;\rm m^3/kg.s^2 {/eq} is the gravitational constant.

The total energy of a satellite is, on the other hand, depends on the same set of parameters, as well on the mass of the satellite by the equation:

{eq}E=\dfrac{-GMm}{2r} {/eq}

  • where {eq}m {/eq} is the mass of the satellite.


{eq}\begin{align*} T&=2\pi\sqrt {\dfrac{r^3}{GM}}\\ \Rightarrow \dfrac{r^3}{GM}&=\dfrac{T^2}{4\pi^2}\\ \Rightarrow r&=\left ( \dfrac{GMT^2}{4\pi^2} \right )^{1/3} \end{align*} {/eq}

After susbtituting {eq}r {/eq} from the above equation into the equation for the energy of a satellite, we have:

{eq}\begin{align*} E&=\dfrac{-GMm}{r}\\ &=\dfrac{-GMm}{\left ( \dfrac{GMT^2}{4\pi^2} \right )^{1/3}}\\ &=-\left ( \dfrac{2\pi GM}{T} \right )^{2/3}m\\ \end{align*} {/eq}

The difference in the energies corresponding to the two time periods is:

{eq}\begin{align*} \Delta E&=-\left ( 2\pi GM\right )^{2/3}m\left ( \dfrac{1}{T_2^{2/3}}-\dfrac{1}{T_1^{2/3}} \right )\\ &=-\left ( 2\pi \times 6.67\times 10^{-11}\times 6\times 10^{24}\right )^{2/3}\times 50\times \left ( \dfrac{1}{\left ( 10.8\times 10^{3} \right )^{2/3}}-\dfrac{1}{\left (7.2\times 10^{3} \right )^{2/3}} \right )\\ &=-\left ( 2.51\times 10^{15}\right )^{2/3}\times 50\times \left ( \dfrac{1}{488}-\dfrac{1}{373} \right )\\ &=-1.36\times 10^{10}\times 50\times \left ( -6.32 \times 10^{-4}\right )\\ &=\boxed{4.3\times 10^{6}\;\rm J} \end{align*} {/eq}

Learn more about this topic:

Stable Orbital Motion of a Satellite: Physics Lab

from Physics: High School

Chapter 8 / Lesson 16

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