is the speed of revolution of Titan (Saturn's moon), when orbital distance is 1.22 ×109 m and time of revolution is 16 days.

7, 091 m s^{- 1}

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

5, 542 m s^{- 1}

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

28, 366 m s^{- 1}

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

14, 183 m s^{- 1}

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

Open in App

Solution

The correct option is B $5, 542 m s^{- 1}$
Given,
Orbital distance, R $= 1.22 \times$ $10^{9} m$
Time period, T $= 16 d a y s = 16 \times 86, 400$ $= 1, 382, 400 s$
We know, speed = $\frac{c i r c u m f e r e n c e}{t i m e p e r i o d}$
= $\frac{2 π R}{T}$
$= \frac{2 \times 3.14 \times 1.22 \times 10^{9}}{1, 382, 400}$
= $5, 542 m s^{- 1}$
$∴$ Speed of Titan with which it revolve around saturn is $5, 542 m s^{- 1}$ .

Suggest Corrections

Similar questions

One of Saturn's moon is named as Mimas. The orbital radius of Mimas is 1.87 $\times$ $10^{8}$ m. The time taken to complete one revolution by Mimas around Saturn is approximately 23 hours. The speed of Mimas with which it revolves around Saturn is

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.

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. In Bohr model of hydrogen atom, the ratio of periods of revolution of an electron in n = 2 and n = 1 orbits is

Q. How many miles away is Saturn's largest moon, Titan from the sun and the earth?

Join BYJU'S Learning Program

is the speed of revolution of Titan (Saturn's moon), when orbital distance is 1.22 ×109 m and time of revolution is 16 days.

The correct option is B 5,542 ms−1Given, Orbital distance, R =1.22× 109 m Time period, T =16 days=16×86,400 =1,382,400 s We know, speed = circumferencetime period = 2πRT =2×3.14×1.22×1091,382,400 = 5,542 ms−1 ∴Speed of Titan with which it revolve around saturn is 5,542 ms−1.