The ratio of escape ve locity at earth (...

The ratio of escape ve locity at earth (`v_e`) to the escape ve locity at a planet (`v_p`) whose radius and mean density are twice as that of earth is

`1:4`

`1:sqrt2`

`1:2`

`1:2sqrt2`

Text Solution

Verified by Experts

Topper's Solved these Questions

FRICTION IS SOLIDS AND LIQUIDS
TARGET PUBLICATION|Exercise MCQ (EVALUATION TEST)|21 Videos
LAWS OF MOTION
TARGET PUBLICATION|Exercise EXERCISE|281 Videos

Similar Questions

Explore conceptually related problems

The ratio of escape velocity at earth (v_(e)) to the escape velocity at a planet (v_(y)) whose radius and density are twice

The escape velocity from the surface of earth is V_(e) . The escape velocity from the surface of a planet whose mass and radius are 3 times those of the earth will be

Time period of second pendulum on a planet, whose mass and diameter are twice that of earth, is

The escape velocity from the surface of the earth of radius R and density rho

v_(e) and v_(p) denotes the escape velocity from the earth and another planet having twice the radius and the same mean density as the earth. Then

Show that the escape velocity of a body from the surface of a planet of radius R and mean density rho is Rfrac(sqrt8pirhoG)(3) OR Show that the escape velocity of a body from the surface of the earth is 2Rfrac(sqrt2pirhoG)(3) , where R is the radius of the earth and rho is the mean density of the earth. OR Obtain formula of escape velocity of a body at rest on the earth's surface in terms of mean density of erth.

Escape velocity of a body from the surface of a spherical planet of mass M, radius R and density p is

The escape velocity of a body from the surface of the earth is V_e and the escape velocity of the body from a satellite orbiting at a height 'h' above the surface of the earth is v_e ' then

The radius of a planet is 1/4 of earth’s radius and its acceleration due to gravity is double that of earth’s acceleration due to gravity. How many times will the escape velocity at the planet’s surface be as compared to its value on earth’s surface

A body weighs 5.6 kgwt on the surface of the Earth. How much will be its the weight on a planet whose mass is 1/7th mass of the Earth and radius twice that of the Earth’s radius.

TARGET PUBLICATION-Gravitation-Exercise

Infinite number of bodies, each of mass 2 kg are situated an x-axis at...
Text Solution
|
The escape velocity from the surface of earth is ve . The escape veloc...
Text Solution
|
The ratio of escape ve locity at earth (ve) to the escape ve locity at...
Text Solution
|
A black hole is an object whose gravitational field is so strong that ...
Text Solution
|
A space station is at a height equal to the radius of the Earth. If VE...
Text Solution
|
A period of geostationary satellite is
Text Solution
|
Two satellites A and B go round a planet in circular orbits having rad...
Text Solution
|
A satellite S is moving in an elliptical orbit around the earth. The m...
Text Solution
|
A remote - sensing satellite of earth revolves in a circular orbit at ...
Text Solution
|
The time period of a geostationary satellite is 24 h, at a height 6RE(...
Text Solution
|
If a satellite has to orbit the earth in a circular path every 6 hrs, ...
Text Solution
|
A satellite is moving with a constant speed 'v' in a circular orbit ab...
Text Solution
|
A satellite is orbiting close to the earth and has a kinetic energy K....
Text Solution
|
A body of mass m is dropped from a height nR above the surface of the ...
Text Solution
|
A satellite has kinetic energy K, potential energy V and total energy ...
Text Solution
|
What is the minimum energy required to launch a satellite of mass m fr...
Text Solution
|
A satellite of mass m is orbiting the earth ( of radius R) at a height...
Text Solution
|
The ratio of binding energy of a satellite at rest on earth's surface ...
Text Solution
|
If the mass of the Sun were ten times smaller and the universal gravit...
Text Solution
|
An extremely small and dense neutron star of mass M and radius R is ro...
Text Solution
|