Escape speed, v=
For another planet both mass and radius are twice that of Earth.
∴v'=
⇒v'==u.
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
vp = (2/3).ve = 0.66 x 11.2
thus,
escape velocity of planet will be
vp = 7.4 km/s
Copyright © 2024 EHUB.TIPS team's - All rights reserved.
Answers & Comments
solution:-
Escape speed, v=![\sqrt{ \frac{2G}{2M} } \sqrt{ \frac{2G}{2M} }](https://tex.z-dn.net/?f=%20%5Csqrt%7B%20%5Cfrac%7B2G%7D%7B2M%7D%20%7D%20)
For another planet both mass and radius are twice that of Earth.
∴v'
=![\sqrt{ \frac{2G\times 2M}{2R} } \sqrt{ \frac{2G\times 2M}{2R} }](https://tex.z-dn.net/?f=%20%5Csqrt%7B%20%5Cfrac%7B2G%5Ctimes%202M%7D%7B2R%7D%20%7D%20)
⇒v'
=
=u.
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