The radius of the nearly circular orbit of mercury is $5.8 \times 10^{10} m$ and its orbital period is 88 days. If a hypothetical planet has an orbital period of 55 days, what is the radius of its circular orbit?

Open in App

Solution

We know that $T^{2} \propto R^{3}$

$⟹ \frac{R_{2}^{3}}{R_{1}^{3}} = \frac{T_{2}^{2}}{T_{1}^{2}}$

$⟹ R_{2}^{3} = \frac{T_{2}^{2}}{T_{1}^{2}} R_{1}^{3}$

$⟹ R^{2} = {(\frac{55}{88})}^{2 / 3} \times 5.8 \times 10^{10} m$

$⟹ R_{2} = 4.24 \times 10^{10} m$

Suggest Corrections

Similar questions

The time period of an artificial satellite in a circular orbit of radius R is 2 days and its orbital velocity is $v_{0}$ . If time period of another satellite in a circular orbit is 16 days then

Q. The time period of an earth satellite in a circular orbit of radius R is 2 days and its orbital velocity is

v_{o}

. If time period of another satellite in a circular orbit is 16 days then

The time period of an artificial satellite in a circular orbit of radius R is 2 days and its orbital velocity is $v_{0}$ . If time period of another satellite in a circular orbit is 16 days then

Q. Two satellites of a planet have period

32

days and

256

days

.

If the radius of the orbit of the former is

R,

the orbital radius of the letter is

Q. A satellite is in a circular orbit around a planet, Its period of revolution is T, radius of the orbit is R, orbital velocity V and acceleration 'a', then:

Join BYJU'S Learning Program

The radius of the nearly circular orbit of mercury is 5.8×1010m and its orbital period is 88 days. If a hypothetical planet has an orbital period of 55 days, what is the radius of its circular orbit?

We know that T2∝R3⟹R32R31=T22T21⟹R32=T22T21R31⟹R2=(5588)2/3×5.8×1010m⟹R2=4.24×1010m

The radius of the nearly circular orbit of mercury is $5.8 \times 10^{10} m$ and its orbital period is 88 days. If a hypothetical planet has an orbital period of 55 days, what is the radius of its circular orbit?

We know that $T^{2} \propto R^{3}$

$⟹ \frac{R_{2}^{3}}{R_{1}^{3}} = \frac{T_{2}^{2}}{T_{1}^{2}}$

$⟹ R_{2}^{3} = \frac{T_{2}^{2}}{T_{1}^{2}} R_{1}^{3}$

$⟹ R^{2} = {(\frac{55}{88})}^{2 / 3} \times 5.8 \times 10^{10} m$

$⟹ R_{2} = 4.24 \times 10^{10} m$