Question

What is magnitude of escape velocity when an object is projected from a plannet whose mass and radius is twice as that of earth

Answer 1

solution:-

Escape speed, v=$\sqrt{ \frac{2G}{2M} }$

For another planet both mass and radius are twice that of Earth.

∴v'$\tiny{e}$=$\sqrt{ \frac{2G\times 2M}{2R} }$

⇒v'$\tiny{e}$=$\sqrt{ \frac{2G}{2M} }$=u.

Answer 2

The escape velocity is given as

v= [2GM/R]1/2

now,

for earth

ve = [2GMe/Re]1/2

or ve = 11.2km/s

for planet

vp = [2GMp/Rp]1/2

and we have

Mp = (1/9)Me

Rp = (1/4)Re

so,

vp = [2G(1/9)Me / (1/4)Re]1/2

or

vp = (2/3).[2GMe / Re]1/2

so,

vp = (2/3).ve = 0.66 x 11.2

thus,

escape velocity of planet will be

vp = 7.4 km/s