Question

a particle is launched tangentially from the surface of the earth with speed v for a particle to move as a satellite which statement is correct​

Answer 1

For the particle to move as a satellite

$\boxed{\frac{v_e}{\sqrt{2}}<v<v_e}$

Therefore, option (2) is correct.

Explanation:

Given:

A particle is launched tangentially from the surface of the earth with speed v

To find out:

The condition for the particle to move as a satellite

Solution:

We know that if the escape velocity of a particle is $v_e$ and its orbital velocity (minimum) is $v_o$ then

$v_e=\sqrt{2}v_o$

or, $v_o=\frac{v_e}{\sqrt{2}}$

Now, if the particle is projected with a velocity v then in order that the particle moves as a satellite, its velocity should be less than the escape velocity $v_e$ while it should be greater than the minimum orbital velocity $v_o$

Thus,

$v_o<v<v_e$

or, $\frac{v_e}{\sqrt{2}}<v<v_e$

Therefore, option (2) is correct.

Hope this answer is helpful.

Know More:

Q: Derive relation between escape velocity and orbital velocity .

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