Gravitational acceleration on the surface of a planet is `(sqrt6)/(11)g.` where g is the gravitational acceleration on the surface of the earth. The average mass density of the planet is `(2)/(3)` times that of the earth. If the escape speed on the surface of the earht is taken to be `11 kms^(-1)` the escape speed on the surface of the planet in `kms^(-1)` will be
The value of escape speed from the surface of earth is
The acceleration due to gravity at the surface of the earth is g . The acceleration due to gravity at a height (1)/(100) times the radius of the earth above the surface is close to :
The ratio of the masses and radii of two planets are 2 : 3 and 4 : 9. What is the ratio of the escape speed at their surface ?
The ratio of the masses and radii of two planets are 2 : 3 and 4 : 9. What is the ratio of the escape speed at their surface ?
Two escape speed from the surface of earth is V_(e) . The escape speed from the surface of a planet whose mass and radius are double that of earth will be.
The ratio of the masses and radii of two planets are 4:6 and 8:18 . What is the ratio of the escape speed at their surface ?
Mass and radius of a planet are two times the value of earth. What is the value of escape velocity from the surface of this planet?
The moon's radius is 1//4 that of the earth and its mass 1//80 times that of the earth. If g represents the acceleration due to gravity on the surface of the earth, that on the surface of the moon is
If the density of the planet is double that of the earth and the radius 1.5 times that of the earth, the acceleration due to gravity on the planet is
Suppose that the acceleration of a free fall at the surface of a distant planet was found to be equal to that at the surface of the earth. If the diameter of the planet were twice the diameter of the earth, then the ratio of mean density of the planet to that of the earth would be
