Question

plz solve this que argent​

Answer 1

Explanation:

1/2mv^2=gmM/r

1/2v^2=gM/r

here it is independent of mass of planet

Answer 2

Question :

Escape velocity of the planet is $\sf{v_e}$. If the radius of the planet remains same and mass becomes 4 times, the escape velocity becomes...

Solution :

❖ The minimum velocity with which a projectile has to be projected to escape the earth's gravitational field is called escape velocity.

Escape velocity of object is given by

$\dag\:\underline{\boxed{\bf{\orange{v_e=\sqrt{\dfrac{2GM}{R}}}}}}$

• The escape velocity doesn't depend on angle of projection from the earth's surface. But as the earth rotates about its axis, so it becomes easier to attain escape velocity if the body is projected in the direction in which the launch site is moving.

In this case radius of the planet remains same so we can say that, $\bf{v_e\propto\sqrt{M}}$

$\sf:\implies\:\dfrac{v_e}{v_e'}=\sqrt{\dfrac{M}{M'}}$

• M' = 4M

$\sf:\implies\:\dfrac{v_e}{v_e'}=\sqrt{\dfrac{1}{4}}$

$\sf:\implies\:\dfrac{v_e}{v_e'}=\dfrac{1}{2}$

$:\implies\:\underline{\boxed{\bf{\red{v_e'=2\times v_e}}}}$

∴ (B) is the correct answer!