Question

) The orbital velocity at a height h above the surface of
earth is 90% of that near the surface of the earth. If
the escape velocity at the surface of earth be Ve then
its value at the height h will be
a) 0.99 Ve.
b) 0.90 Ve.
c) 0.81 Ve.
d) 0.11Ve.​

Answer 1

Answer:

option A 0.99 Ve

Explanation:

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Answer 2

Answer:

The escape velocity at height, h is 0.9 times of escape velocity at surface of Earth.

Explanation:

Given that :

• Orbital velocity at h = 90% of orbital velocity at surface.

Solution :

• Orbital velocity of a body at a distance r from the centre of Earth is

$V₀ = \sqrt{ \frac{GM}{r} }$

• Escape velocity of a body at a distance r from centre of Earth is

$Ve = \sqrt{ \frac{2GM}{r} }$

• Let, the mass of Earth be M and radius of Earth be R.
• It is given that, Orbital velocity at height, h above the surface of Earth is equal to 90% of orbital velocity at surface.

$\sqrt{ \frac{GM}{ R+ h} } = \frac{90}{100} \sqrt{ \frac{GM}{R} } \\ squaring \: both \: sides \\ \frac{GM}{ R+ h} =0.81 \frac{GM}{ R} \\ R = 0.81(R+ h) \\ 0.19R = 0.81 h \\ h = \frac{19}{81} R \: \: - - (1)$

• Now, Let the escape velocity at surface be Ve and at distance h above the surface of Earth be Ve'. Then,

$Ve'= \sqrt{ \frac{2GM}{R + h} } \\ using \: value \: of \: h \: from \: equation(1) \\ = \sqrt{ \frac{2GM}{R + \frac{19}{81}R } } \\ = \sqrt{ \frac{81}{100} \frac{2GM}{R} } \\ = \frac{9}{10} \sqrt{ \frac{2GM}{R } } = 0.9Ve$

• Hence, escape velocity at height h is 0.9 times of Ve .