The escape speed from earth's surface is `11kms^(-1)`. A certain planet has a radius twice that of earth but its mean density is the same as that of the earth. Find the value of the escape speed from the planet.
Text Solution
Verified by Experts
Let `M,R,rho` be the mass radius and density of earth and `M,R,rho` be the corresponding vaues of the planet. Then escape speed from earth's surface is `v_(e)=sqrt((2GM)/R)=sqrt((2G)/Rx4/3piR^(3))rho=sqrt((8piGrhoR^(2))/(3))` Escape speed from planet's surface is `v_(e)=sqrt((8piGrho(2R)^(2))/3)=2sqrt((8piGrhoR^(2))/3)=2v_(e)=2xx11=22km//s`
Topper's Solved these Questions
GRAVITATION
CENGAGE PHYSICS|Exercise Exercise 6.1|13 Videos
GRAVITATION
CENGAGE PHYSICS|Exercise Exercise 6.2|15 Videos
FLUID MECHANICS
CENGAGE PHYSICS|Exercise INTEGER_TYPE|1 Videos
KINEMATICS-1
CENGAGE PHYSICS|Exercise Integer|9 Videos
Similar Questions
Explore conceptually related problems
The value of escape speed from the surface of earth is
A planet has twice the radius but the mean density is 1/4 th as compared to earth. What is the ratio of escape velocity from earth to that from the planet
What will be the escape speed from a planet having radius thrice that of earth and the same mean density as that of the earth ? (Take v_(e)=11.2 km/s)
The escape velocity form the earth is 11 kms^(-1) the esacpe velocity from a planet having twice the radius and the same mean denisty as the earth would be
The escape speed from jupiter is approximately 59.5 kms^(-1) and its radius is about 12 times that to earth. From this we may estimate the mean density of jupiter to be about (radius of earth =escape speed from the earth is 11.2 kgm^(-1)
The escape velocity of the earth is 11.2 km/s. For a planet whose mass and radius are twice those of the earth, the escape velocity will be
