The escape velocity from the Earth's surface is . The escape velocity from the surface of another planet having a radius, four times that of Earth and same mass density is [2021]

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

More

Solution

Q.
The escape velocity from the Earth's surface is $v$ . The escape velocity from the surface of another planet having a radius, four times that of Earth and same mass density is [2021]

1 $4 v$

2 $v$

3 $2 v$

4 $3 v$

Ans.
(1)

The formula of escape velocity is

$v = \sqrt{\frac{2 G M}{R}}$

where, $G$ = gravitational constant, $M$ = mass and $R$ = radius.

Mass, $M = \frac{4}{3} π R^{3} ρ$ where, $ρ$ is the density of the planet

   $v = \sqrt{\frac{2 \times G}{R} \times \frac{4}{3} π R^{3} ρ}$

$v = \sqrt{\frac{8 G π}{3} R^{2} ρ}$ ...(i)

Let the escape velocity on planet is $v'$ .

For planet, $R' = 4 R, ρ' = ρ$

$v' = \sqrt{\frac{8 G π}{3} R'^{2} ρ'}$ ...(ii)

Divide equation (i) by (ii), we get

So,   $\frac{v'}{v} = \sqrt{\frac{R'^{2}}{R^{2}} \times \frac{ρ'}{ρ}}$

   $\frac{v'}{v} = \sqrt{\frac{16 R^{2}}{R^{2}} \times \frac{ρ}{ρ}} = \sqrt{16} = 4$

   $v' = 4 v$

Want Us to Email you About Special Offers & Updates?