Question

An asteroid, whose mass is 4.20×10⁻⁴ times the mass of Earth, revolves in a circular orbit around the Sun at a distance that is 6 times the Earth's distance from the Sun. Calculate the period of revolution of the asteroid.

Answer 1

Answer:

The time period of revolution of the asteroid is 14.69 years.

Explanation:

Given that,

Mass of asteroid $M= 4.20\times10^{-4}M_{e}$

Distance $r= 6r_{e}$

We need to calculate the velocity

Using relation centripetal force and gravitational force

$\dfrac{mv^2}{r}=\dfrac{GMm}{r^2}$

$v^2=\sqrt{\dfrac{GM}{r}}$

We need to calculate the time period of revolution of the asteroid

Using formula of time period

$T=\dfrac{2\pi r}{v}$

Put the value of v into the formula

$T=\dfrac{2\pi r}{\sqrt{\dfrac{GM}{r}}}$...(I)

We need to calculate the time period of revolution of the earth

Using formula of time period

$T_{e}=\dfrac{2\pi r}{v}$

$T_{e}=\dfrac{2\pi r_{e}}{\sqrt{\dfrac{GM_{e}}{r_{e}}}}$....(II)

From equation (I) and (II)

$\dfrac{T^2}{T_{e}^2}=(\dfrac{r}{r_{3}})^3$

$\dfrac{T^2}{T_{e}^2}=216$

$\dfrac{T}{T_{e}}=\sqrt{216}$

$\dfrac{T}{T_{e}}=14.69$

$T=14.69T_{e}$

Hence, The time period of revolution of the asteroid is 14.69 years.