Four point mass bodies of masses as shown in Fig. are placed at the vertices of a square `ABCD`, gravitational force on the body at `A`. Given, `G = 6.6 xx 10^(-11) Nm^(2) kg^(-2)`

Universal gravitational constant (a) G = 6.684 xx 10^(-10) Nm^(2) kg^(-1)

Two spheres of 1kg mass each are separated by 3 m. Calculate the gravitational force between then. Given G = 6.67 xx 10^ -11Nm^2//kg^2 .

A point mass body of mass 2 kg is placed at a distance 20 cm from one end of a uniform rod of length 2 m and mass 10 kg . Calculate (i) gravitational intensity at the location of point mass body due to the rod. (ii) gravitational force on the body due to the rod. Use G = 6.67 xx 10^(-11) Nm^(2) kg^(-2) .

If the value of universal gravitational constant in SI is 6.6 xx 10^(-11) Nm^(2) kg^(-2) , then find its value in CGS system.

Calculate the force of attraction between two balls each of mass 10kg when their centres are 10 cm apart. The value of gravitational constant G = 6.67 xx 10^(-11) Nm^(2) kg^(-2)

The mass of the earth is 6.0 xx 10^(24) kg and its radius is 6.4 xx 10^(6) m . How much work will be done in taking a 10 kg body from the surface of the Earth to infinity ? What will be the gravitational potential energy of the body on the Earth's surafce ? G = 6.67 xx 10^(-11) Nm^(2) kg^(-2) .

Two bodies of masses 100 kg and 10,000 kg are at a distance 1 m apart. At which point on the line joining them will the resultant gravitational field intensity is zero? What is the gravitational potential at that point ? G = 6.67 xx 10^(-11) Nm^(2) kg^(-2) .

The boides of masses 100 kg and 8100 kg are held at a distance of 1 m. The gravitational field at a point on the line joining at them is zero. The gravitational potential at the point in J/kg is (G = 6.67 xx 10^(-11) Nm^(2 // kg^(2))