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

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

Step-by-step explanation:

Given that,

Mass of asteroid
`M= 4.20*10^(-4)M_(e)`

Distance
`r= 6r_(e)`

We need to calculate the velocity

Using relation centripetal force and gravitational force

`(mv^2)/(r)=(GMm)/(r^2)`

`v^2=\sqrt{(GM)/(r)}`

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

Using formula of time period

`T=(2\pi r)/(v)`

Put the value of v into the formula

`T=\frac{2\pi r}{\sqrt{(GM)/(r)}}`...(I)

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

Using formula of time period

`T_(e)=(2\pi r)/(v)`

`T_(e)=\frac{2\pi r_(e)}{\sqrt{(GM_(e))/(r_(e))}}`....(II)

From equation (I) and (II)

`(T^2)/(T_(e)^2)=((r)/(r_(3)))^3`

`(T^2)/(T_(e)^2)=216`

`(T)/(T_(e))=√(216)`

`(T)/(T_(e))=14.69`

`T=14.69T_(e)`

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