Question

The minimum increase in orbital
velocity required for a satellite
revolving around the earth in a
circular orbit so that it can just escape
from the gravitational field of earth is
1) 50% 2) 100%
3) 41.4 % 4) 73.1%

Answer 1

Heyaa friend ! ☺

>> Here's your answer :

• The minimum increase in orbital velocity required for a satellite revolving around the earth in a circular orbit so that it can just escape from the gravitational field of earth is:

3) 41.4 %

_________________________

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Answer 2

Answer:

The minimum increase in orbital velocity required for a satellite revolving around the earth in a circular orbit so that it can just escape from the gravitational field of the earth is 41.4 %.

Explanation:

Assumption,

Let the velocity with which the satellite is revolving around the Earth be 'v'.

Let the velocity required for leaving the Earth's orbit be 'v1'.

Given,

The satellite is revolving around the Earth.

To find,

The velocity required for leaving the Earth's orbit (v1).

Calculation,

The force of gravity of the satellite is balanced by the centrifugal force of the satellite.

i.e.

$\frac{mv^2}{R+h} = \frac{GmM}{(R + h)^2}$ (let the mass and height from the surface of the Earth be m and h).

$v = \sqrt{\frac{GM}{(R + h)} } \approx \sqrt{\frac{GM}{R} } = \sqrt{gR}$

(As h can be neglected compared to the radius of the Earth)

Now for the satellite to escape from its orbit it must increase its speed to v1.

So, by conservation of energy

$\frac{1}{2} m(v1)^2= \frac{GmM}{R}$

$\implies v1 = \sqrt{\frac{2GM}{R} }= \sqrt{2gR}$

hence, the increase in speed is v1 - v

= √(2gR) - √(gR)

= √gR (√2 - 1)

So the percentage in the increase is:

$\frac{v1}{v} \times 100$

$\implies \frac{\sqrt{gR} (\sqrt{2} - 1) }{\sqrt{gR} } \times 100$

⇒ (√2 - 1) × 100

⇒ 41.4 %

Therefore, the minimum increase in orbital velocity required for a satellite revolving around the earth in a circular orbit so that it can just escape from the gravitational field of the earth is 41.4 %.

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