Approximately one billion years ago, the Moon orbited the Earth much closer than it does today. The radius of the orbit was only 24400 km. Today, the radius is 385000 km. The orbital period was only 23400 s. The present period is $2.36 \times 10^{6} s$ . Assume that the orbit of the Moon is circular. Calculate the ratio of the speed of the Moon in its ancient orbit to the speed that it has today.

15.8

No worries! We‘ve got your back. Try BYJU‘S free classes today!

12.8

No worries! We‘ve got your back. Try BYJU‘S free classes today!

10.2

No worries! We‘ve got your back. Try BYJU‘S free classes today!

6.39

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D
6.39

Linear velocity in a circular orbit is given by:

$v = \frac{2 \times π \times r}{t}$

$\frac{v_{1}}{v_{2}} = \frac{r_{1}}{r_{2}} \times \frac{t_{2}}{t_{1}} = \frac{24400}{385000} \times \frac{2.36 \times 10^{6}}{23400} = 6.39$

Suggest Corrections

Similar questions

Q. One of saturn's moon is named as Mimas. The orbital radius of Mimas is

1.87 \times

10^{8} m

from saturn. The mean orbital period of Mimas is approximately 23 hours. Calculate the speed of Mimas with which it revolves around saturn. Assume that the path taken is circular.

Q. One of Saturn’s moons is named Mimas. The orbital radius of Mimas is 1.87 x

10^{8}

m from Saturn. The mean orbital period of Mimas is approximately 23 hours. Its speed with which it revolves around Saturn is ________. (Assume that the path taken is circular.)

Q. One of Saturn's moon is named as Mimas. The orbital distance of Mimas is

1.87 \times 10^{8} m

from Saturn. The mean orbital period of Mimas is approximately 23 hours. Calculate the speed of Mimas with which it revolves around Saturn. Assume the path of the moon to be perfectly circular.

Q. The orbital radius of moon around the earth is

3.8 \times 10^{8}

meter and its time period is 27.3 days. The centripetal acceleration of moon will be

Mimas is one of Saturn's many moons.. The orbital distance of Mimas, from Saturn is 1.87 $\times$ $10^{8}$ m. Its mean orbital period is approximately 23 hours. Calculate the speed with which it revolves around Saturn.

Join BYJU'S Learning Program

The correct option is D 6.39 Linear velocity in a circular orbit is given by: v=2×π×rt v1v2=r1r2×t2t1=24400385000×2.36×10623400=6.39

The correct option is D
6.39

Linear velocity in a circular orbit is given by:

$v = \frac{2 \times π \times r}{t}$

$\frac{v_{1}}{v_{2}} = \frac{r_{1}}{r_{2}} \times \frac{t_{2}}{t_{1}} = \frac{24400}{385000} \times \frac{2.36 \times 10^{6}}{23400} = 6.39$