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

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.

What is the maximum height attained by a body projected with a velocity equal to one- third of the escape velocity from the surface of the earth? (Radius of the earth=R)

What is the maximum height attained by a body projected with a velocity equal to one- third of the escape velocity from the surface of the earth? (Radius of the earth=R)

A ball is thrown vertically upwards with a velocity equal to half the escape velocity from the surface of the earth. The ball rises to a height h above the surface of the earth. If the radius of the earth is R, then the ratio h/R is

The escape velocity from the surface of the earth is (where R_(E) is the radius of the earth )

The escape velocity from the surface of the earth is (where R_(E) is the radius of the earth )

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

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

The escape velocity of a body from the surface of the earth is V_(1) and from an altitude equal to twice the radius of the earth, is, V_(2) . Then