If the radius of moon is 1-7 - 106 m and its mass is 7-34- 1022 kg- Then its escape velocity is

The correct option is A 2.4× 10^3 ms^ 1 V_e=√(2GM/R) =√(2(6.67×10^ 11)(7.34×10^22)/1.7×10^6) =2.4× 10^3 m/s ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Escape Speed

If the radius...

Question

If the radius of moon is $1.7 \times 10^{6} m$ and its mass is $7.34 \times 10^{22} k g$ . Then its escape velocity is

2.4 \times 10^{3} m s^{- 1}

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

2.4 \times 10^{2} m s^{- 1}

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

3.4 \times 10^{3} m s^{- 1}

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

3.4 \times 10^{2} m s^{- 1}

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

Open in App

Solution

The correct option is A $2.4 \times 10^{3} m s^{- 1}$
$V_{e} = \sqrt{\frac{2 G M}{R}}$
$= \sqrt{\frac{2 (6.67 \times 10^{- 11}) (7.34 \times 10^{22})}{1.7 \times 10^{6}}}$
$= 2.4 \times 10^{3}$ m/s

Suggest Corrections

Similar questions

Q. The kinetic energy of an electron is

4.55 \times 10^{- 25} J

.The mass of electron is

9.1 x 10^{- 31} kg

.The velocity of the electron is .

The kinetic energy of an electron is $4.55 \times 10^{- 25} J$ . The mass ofelectron is $9.1 \times 10^{- 31} k g$ . Calculate velocity of the electron.

Q. A photon with initial frequency

10^{11}

Hz scatters off an electron at rest. Its final frequency is 0.9 X

10^{11}

Hz. The speed of scattered electron is close to

(h = 6.63 \times 10^{- 34}

J s,

m_{e} = 9.1 \times 10^{- 31}

kg)

Q. Find the escape velocity on the surface of the moon, if the mass and radius of the moon are

7.34 \times 10^{22} k g

and

1.74 \times 10^{6} m

respectively.

Calculate the escape velocity on the surface of the moon, if the mass and radius of the moon are $7.34 \times 10^{22} k g$ and $1.74 \times 10^{6} m$ respectively.

Join BYJU'S Learning Program

If the radius of moon is 1.7×106 m and its mass is 7.34×1022 kg. Then its escape velocity is

The correct option is A 2.4×103 ms−1Ve=√2GMR=√2(6.67×10−11)(7.34×1022)1.7×106=2.4×103 m/s

If the radius of moon is $1.7 \times 10^{6} m$ and its mass is $7.34 \times 10^{22} k g$ . Then its escape velocity is

The correct option is A $2.4 \times 10^{3} m s^{- 1}$
$V_{e} = \sqrt{\frac{2 G M}{R}}$
$= \sqrt{\frac{2 (6.67 \times 10^{- 11}) (7.34 \times 10^{22})}{1.7 \times 10^{6}}}$
$= 2.4 \times 10^{3}$ m/s