Question

SV
2
7
Escape velocity of an atmospheric particle which is 1000 km above the earth's surface, is (radius of earth is
6400 km and g = 9.8 m/s2)
(1) 6.5 km/s
(2) 8 km/s
(3)
10 km/s
(4) 11.2 km/s
pls don't irrelevant answers , if you dont know the answer then just ignore the question​

Answer 1

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Answer 2

Answer:

c. 10 km/s

Explanation:

Hint: The escape velocity is a velocity that is required by an object in order to overcome the gravitational pull of a body (like earth). The kinetic energy of the object has to be more than (or equal to) the gravitational potential of the earth.

Formula used:

Escape velocity is given by

v=√2GMr

r is the distance from the centre of the earth.

Complete answer:

Consider an object at a height of h from the earth surface. This means that it is at a distance of r=R+h from the center of the earth.

The escape velocity is determined by equating the kinetic energy of the particle with the gravitational potential it experiences due to earth. If the kinetic energy is more than the gravitational pull, then certainly the object will escape the earth's field.

We know that the gravitational potential has a magnitude given by:

GMmr

And kinetic energy as usual is

12mv2

So, equating the two gives us:

12mv2=GMmr

This gives us the escape velocity:

v=√2GMr

We are given the value of g and we know g is expressed as:

g=GMR2

Therefore we modify the escape velocity formula as:

v=2GMR2rR2−−−−−−−√

which upon keeping g, can be simply written as:

v=2gR2r−−−−−√

Keeping r=R+h and substituting the given values i.e., h = 1000 km, R = 6400 km and g = 9.8 m/s2;

v=2×9.8×(6400000)27400000−−−−−−−−−−−−−−−−−√

This gives the velocity to be about

v= 10415.79 m/s

Therefore the correct answer is 10.4 km/s.

So, the correct answer is “Option C”.