Tethys, one of Saturn's moons, travels in a circular orbit at a speed of 1.1x104 m/s. The mass of...

Question:

Tethys, one of Saturn's moons, travels in a circular orbit at a speed of 1.1x104 m/s. The mass of Saturn is 5.67x1026 kg. Calculate

a) the orbital radius in kilometers.

b) the orbital period in Earth days.

Speed of a Satellite:

{eq}\\ {/eq}

A satellite is an object that orbits around a celestial body like planets, stars, moons, etc. In the case of circular orbits, the centripetal force for the circular motion of the satellite is provided by the gravitational force between the planet and the satellite.

Since the speed of an object in uniform circular motion is related to the centripetal force acting on it, and the radius of the motion, the speed of a satellite in a circular orbit is related to the gravitational force, which in turn depends on the mass of the planet, and the distance of the satellite from the planet.

Answer and Explanation:

{eq}\\ {/eq}

We are given:

  • The speed of Tethys, {eq}v=1.1\times 10^{4}\;\rm m/s {/eq}
  • The mass of the Saturn, {eq}M=5.67\times 10^{26}\;\rm kg {/eq}

a)

The 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*} 2.2\times 10^{4}&=\sqrt{\dfrac{6.67\times 10^{-11}\times 5.67\times 10^{26}}{r}}\\ \Rightarrow \dfrac{37.82\times 10^{15}}{r}&=4.84\times 10^{8}\\ \Rightarrow r&=\dfrac{37.82\times 10^{15}}{4.84\times 10^{8}}\\ &=\boxed{7.81\times 10^{7}\;\rm m}\\ &=\boxed{7.81\times 10^{4}\;\rm km} \end{align*} {/eq}


b)

Since the orbit is circular, the distance covered in an orbit is:

{eq}\Delta s=2\pi r=2\pi\times 7.81\times 10^{7}=4.91\times 10^{8}\;\rm m {/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}

Therefore, the period of the satellite is:

{eq}\begin{align*} \Delta t&=\dfrac{\Delta s}{v}\\ &=\dfrac{4.91\times 10^{8}}{1.1\times 10^{4}}\\ &=\boxed{4.46\times 10^{4}\;\rm s}\\ &=4.46\times 10^{4}\times \dfrac{1}{3600\times 24} \;\rm days\\ &=\boxed{0.516\;\rm days} \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
565

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