A bullet is fired vertically upwards with velocity v from the surface of a spherical planet. When it reaches its maximum height, its acceleration due to the planet’s gravity is $${\left( {{1 \over 4}} \right)^{th}}$$ of its value at the surface of the planet. If the escape velocity from the planet is $${v_{esc}} = v\sqrt N $$, then the value of N is (ignore energy loss due to atmosphere)

Gravitational acceleration on the surface of a planet is $${{\sqrt 6 } \over {11}}g$$, where $$g$$ is the gravitational acceleration on the surface of the earth. The average mass density of the planet is $${2 \over 3}$$ times that of the earth. If the escape speed on the surface of the earth is taken to be 11 kms^-1, the escape speed on the surface of the planet in kms^-1 will be

A binary star consists of two stars A (mass 2.2M_s) and B (mass 11M_s), where M_s is the mass of the sun. They are separated by distance d and are rotating about their centre of mass, which is stationary. The ratio of the total angular momentum of the binary star to the angular momentum of star B about the centre of mass is