Question

g Two planets are in circular orbits around a star of with unknown mass. One planet is orbiting at a distance r1=150×106 km and has an orbital period T1=240 days. The second planet has an orbital radius r2=230×106 km. Find the orbital period of the second planet T2.

Answer 1

456 days.

Step-by-step explanation:

Given,

r₁ = 150 x 10⁶ Km

r₂ = 230 x 10⁶ Km

T₁ = 240 days

T₂ = ?

Using Kepler's law

`T^2\ \alpha \ r^3`

Now,

`(T_2^2)/(T_1^2)=(r_2^3)/(r_1^3)`

`T_2=\sqrt{T_1^2* (r_2^3)/(r_1^3)}`

`T_2=\sqrt{240^2* ((230* 10^6)^3)/((150*  10^6)^3)}`

`T_2 = 455.68\ days`

Time taken by the second planer is equal to 456 days.

Answer 2

Time period of second planet will be 126.40 days

Step-by-step explanation:

We have given radius of first planet
`r_1=150* 10^6km=150* 10^9m`

Orbital speed of first planet
`T_1=240days`

Radius of second planet
`r_2=230* 10^6km=230* 10^9m`

We have to find orbital period of second planet

Period of orbital is equal to
`T=2\pi \sqrt{(r^3)/(G(M_1+M_2))}`

From the relation we can see that
`T=r^{(3)/(2)}`

`(T_1)/(T_2)=((r_1)/(r_2))^(3)/(2)`

`(240)/(T_2)=((150* 10^9)/(230* 10^9))^(3)/(2)`

`T_2=126.40 days`

Time period of second planet will be 126.40 days