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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 :

Answer:

Time period of second planet will be 126.40 days

Explanation:

We have given radius of first planet [tex]r_1=150\times 10^6km=150\times 10^9m[/tex]

Orbital speed of first planet [tex]T_1=240days[/tex]

Radius of second planet [tex]r_2=230\times 10^6km=230\times 10^9m[/tex]

We have to find orbital period of second planet

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

From the relation we can see that [tex]T=r^{\frac{3}{2}}[/tex]

[tex]\frac{T_1}{T_2}=(\frac{r_1}{r_2})^\frac{3}{2}[/tex]

[tex]\frac{240}{T_2}=(\frac{150\times 10^9}{230\times 10^9})^\frac{3}{2}[/tex]

[tex]T_2=126.40 days[/tex]

Time period of second planet will be 126.40 days

Answer:

456 days.

Explanation:

Given,

r₁ = 150 x 10⁶ Km

r₂ = 230 x 10⁶ Km

T₁ = 240 days

T₂ = ?

Using Kepler's law

[tex]T^2\ \alpha \ r^3[/tex]

Now,

[tex]\dfrac{T_2^2}{T_1^2}=\dfrac{r_2^3}{r_1^3}[/tex]

[tex]T_2=\sqrt{T_1^2\times \dfrac{r_2^3}{r_1^3}}[/tex]

[tex]T_2=\sqrt{240^2\times \dfrac{(230\times 10^6)^3}{(150\times  10^6)^3}}[/tex]

[tex]T_2 = 455.68\ days[/tex]

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

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