Question

calculate the escape velocity of the body from the moon .whose radius is 1.76 ×10^6 m ans mass of moon = 7.36 × 10^22kg​

Answer 1

Answer:

23.6 m/s

Explanation:

Given,

Radius (r) = 1.76*10^6 m

Mass (m) = 7.36*10^22 kg

G = 6.67*10^(-11) Nm^2/kg^2

We know,

Escape velocity (v) = √(2Gm/r)

Therefore,

$v = \sqrt{ \frac{2 \times 6.67 \times {10}^{ - 11} \times 7.36 \times {10}^{22} }{1.76 \times {10}^{6} } } \\ v = 23.6 \times {10}^{2} m {s}^{ - 1}$

HOPE IT HELPS

Answer 2

Answer:

your answer is below,

well it was above my head to know this

Explanation:

Escape velocity from the surface of the moon is,

$V_{e} \sqrt{\frac{2GM}{R} }$

where,  

M is the mass of the moon, and  

R is radius of the moon.

Here, we have

$R= 1.74$ × $10^6$

$M=7.36$ × $10^2^2$

$G= 6.67$ × $10^-^1^1Nm^2/kg^2$

∴ Escape velocity is,

$Ve = \sqrt{\frac{2 \ x \ 6.67 \ x \ 10^-^1^1 \ x \ 7.36 \ x \ 10^2}{1.74 \ x\ 10^6} } m/s$

$= 2.3754 \ x\ 10^3\ m/sec$

$= 2.38 \ km/sec$

Since escape velocity for the moon is 2.38 km/sec which is small as compared to the average speed of gas molecules at the moon, therefore the moon cannot bind the atmosphere.