A planet with a mass of (7x10^{21}) kg is in a circular orbit around a star with a mass of (2 x...

Question:

A planet with a mass of 7 x {eq}10^{21} {/eq} kg is in a circular orbit around a star with a mass of 2 x {eq}10^{30} {/eq} kg. The planet has an orbital radius of 3 x {eq}10^{10} {/eq} m.

a) What is the linear orbital velocity of the planet?

b) What is the period of the planet's orbit?

c) What is the total mechanical energy of the planet?

Orbital Motion of a Satellite:

{eq}\\ {/eq}

Satellites are the class of objects that orbit around a planet/star/moon etc. The centripetal force for the orbital motion of the satellite is provided by the gravitational force of attraction between the satellite and the planet.

The period, as well as the velocity of a satellite, assuming a circular orbit, depends on the mass of the planet/star/moon orbited by the satellite, and the distance between the satellite from the center of the planet.

The energy of a satellite consists of its kinetic energy and gravitational potential energy, both of which depends on the previous factors, as well as the mass fo the satellite.

Answer and Explanation:

{eq}\\ {/eq}

We are given:

  • The mass of the planet, {eq}m=7\times 10^{21}\;\rm kg {/eq}
  • The mass of the star, {eq}M=2\times 10^{30}\;\rm kg {/eq}
  • The orbital radius, {eq}r=3\times 10^{10}\;\rm m {/eq}


a)

The linear orbital speed of a satellite depends only on the mass of the planet/star, {eq}M {/eq}, orbited by the satellite, and the radius, {eq}r {/eq}, of the circular orbit followed by the satellite by the following equation:

{eq}v=\sqrt{\dfrac{GM}{r}} {/eq}

Here,

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


After plugging the given values into the above equation, we have:

{eq}\begin{align*} v&=\sqrt{\dfrac{6.67\times 10^{-11}\times 2\times 10^{30}}{3\times 10^{10}}}\\ &=\sqrt{44.47\times 10^{8}}\\ &=\boxed{6.67\times 10^{4}\;\rm m/s} \end{align*} {/eq}


b)

Similarly, 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}


After plugging the given values into the above equation, we have:

{eq}\begin{align*} T&=2\pi\sqrt {\dfrac{\left ( 3\times 10^{10} \right )^3}{6.67\times 10^{-11}\times 2\times 10^{30}}}\\ &=2\pi\sqrt {20.24\times 10^{10}\;\rm s^2}\\ &\approx\boxed{4.5\times 10^{5} \;\rm s} \end{align*} {/eq}


c)

The total mechanical energy of a satellite depends only on the mass of the satellite, {eq}m {/eq}, the mass of the planet/star, {eq}M {/eq}, orbited by the satellite, and the radius, {eq}r {/eq}, of the circular orbit followed by the satellite by the following equation:

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

Plugging in the given values, we have:

{eq}\begin{align*} E&=-\dfrac{6.67\times 10^{-11}\times 2\times 10^{30}\times 7\times 10^{21}}{2\times 3\times 10^{10}}\\ &=\boxed{-1.56\times 10^{31}\;\rm J} \end{align*} {/eq}



Learn more about this topic:

Loading...
Stable Orbital Motion of a Satellite: Physics Lab

from Physics: High School

Chapter 8 / Lesson 16
627

Related to this Question

Explore our homework questions and answers library