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A planet orbits the sun at a distance of 2.87 x 10^9 km. If the mass of the sun is 1.99 x 10^30...

Question:

A planet orbits the sun at a distance of {eq}2.87*10^9 \ km {/eq}. If the mass of the sun is {eq}1.99 *10^{30} \ kg {/eq},

(a) Find the orbital period of the planet.

(b) Calculate the orbital speed of the planet.

Speed and Period of a Satellite:

{eq}\\ {/eq}

A satellite performs orbital motion around its parent celestial body (star/planet/moon, etc) by the virtue of the centripetal force provided by the means of the gravitational force between the satellite and the parent celestial body.

In the case of circular orbits around the parent body, the velocity of the satellite remains constant and depends on the following factors:

  • The mass of the parent celestial body.
  • The distance of the satellite from the parent celestial body.

The time period of a satellite is defined as the time required by the satellite to make 1 complete orbit. Once the velocity of the satellite is known, the time period can be calculated by simply dividing the total distance covered in an orbit (circumference of the circular orbit) by the speed of the satellite.

Answer and Explanation:

{eq}\\ {/eq}

We are given:

  • The radius of the orbital motion fo the planet, {eq}r=2.87\times 10^{9}\;\rm km=2.87\times 10^{12}\;\rm m {/eq}
  • The mass of the sun, {eq}M=1.99\times 10^{30}\;\rm kg {/eq}


b)

The speed of a satellite depends only on the mass of the planet/star orbited by the satellite and the radius of the orbit followed by the satellite:

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

Here,

  • {eq}M {/eq} is the mass of the planet/star orbited by the satellite.
  • {eq}r {/eq} is the radius of orbit.
  • {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 1.99\times 10^{30}\;\rm m^3\cdot kg^{-1}\cdot s^{-2}\cdot kg}{2.87\times 10^{12}\;\rm m}}\\ &=\sqrt{46.25\times 10^{6}\;\rm m^2\cdot s^2}\\ &=\boxed{6.8\times 10^{3}\;\rm m/s} \end{align*} {/eq}


a)

The distance traveled by the satellite in one orbit is:

{eq}\Delta s=2\pi r {/eq}

The speed of an object is equal to the distance covered by the object per unit time. That is:

{eq}v=\dfrac{\Delta s}{\Delta t} {/eq}

Here,

  • {eq}\Delta s {/eq} is the distance travelled in time {eq}\Delta t {/eq}

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

{eq}\begin{align*} v&=\dfrac{2\pi r}{\Delta t}\\ \Rightarrow \Delta t&=\dfrac{2\pi r}{v}\\ &=\dfrac{2\pi r}{\sqrt{GM/r}}\\ &=2\pi\sqrt {\dfrac{r^3}{GM}}\\ &=2\pi\sqrt {\dfrac{\left (2.87\times 10^{12} \right )^3\;\rm m^3}{6.67\times 10^{-11}\times 1.99\times 10^{30}\;\rm m^3\cdot kg^{-1}\cdot s^{-2}\cdot kg}}\\ &=2\pi\sqrt{17.81\times 10^{16}\;\rm s^2}\\ &=\boxed{2.65\times 10^{9}\;\rm s} \end{align*} {/eq}



Learn more about this topic:

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Stable Orbital Motion of a Satellite: Physics Lab

from Physics: High School

Chapter 8 / Lesson 16
548

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