Question

a rocket is fired with a speed v = 2√gr near the earth's surface and direction upwards.
(a) show that it will escape from the earth.
(b) show that in interstellar space it's speed is v = √2gR​

Answer 1

Answer:

36

Explanation:

Answer 2

AnSwer :-

(a) As PE of rocket at the surface of the earth is (-GMm/R) and at infinity is zero, energy required for escaping from earth

$\sf = 0 - \bigg( \dfrac{GMm}{R} \bigg) = mgR \: \: \: \: \: \: \bigg[ \because g = \dfrac{GM}{{R}^{2} } \bigg]$

And as initial KE of the rocket

$\sf \dfrac{1}{2} m {v}^{2} = 2mgR$

is greater than the energy required for escaping (= mgR), the rocket will escape.

(b) If v is the velocity of the rocket in interstellar space (free from gravitational effects) then by Conversion of energy,

$\sf \dfrac{1}{2}m \Big(2 \sqrt{gR} \Big)^{2} - \dfrac{1}{2} m \Big( \sqrt{2gR} \Big)^{2} = \dfrac{1}{2} {mv}^{2}$

$\sf {v}^{2} = 4gR - 2gR$

$\sf v = \sqrt{2gR}$