Calculate the acceleration due to gravity on the earth from the following data
`G= 6.6 xx 10 ^(-11) N m^(2) Kgm^(-3)`
Mean density of the earth =`5xx 10 ^(-3) kgm`
Radius of the earth =`6.4 xx10 ^(6) m`.

Calculate the acceleration due to gravity on the earth 's surface , given that gravitational constant G = 6.7 xx 10^(-11) SL unit , radius of he earth = 6. 4 xx 10 ^(6) m and mean density of earth = 5.5 xx 10^(3) kg m^(-3) .

With what velocity must a body be projected vertically upwards from the surface of the earth in orer that it may never return to the earth ? ( G = 6.66 10 ^(-11) SI units , mean density of the earth = 5.525 xx 10^(3) kg m^(-3) , radius of the earth = 6.38 xx 10 ^(6) m.)

Calculate the value of acceleration due to gravity at a place of latitude 45^(@) . Radius of the earth = 6.38 xx 10^(3) km .

Calculate the speed of projection necessary to send a body right out of the filed of the earth's grvivtational attraction. G= 6.63 xx10^(-11) "newton metre^(2) " Kg^(2) Mass of the earth = 5.97 xx 10^(24) Radius of the earth = 6.37 xx 10^(6) m .

Calculate the mass and density of the earth. Given that Gravitational constant G = 6.67 xx 10^(-11) Nm^(2)// kg^(2) , the radius of the earth = 6.37 xx 10^(6) m and g = 9.8 m//s^(2) .

Find the magnitude of the acceleration due to gravity at the surface of the earth. ( M = 6 xx 10^(24) Kg, R = 6400 Km)

A mango of mass 0.3 kg falls from a tree. Calculate the acceleration of the mango towards the earth. Also calculate the acceleration of the earth towards the mango. Take, Mass of the earth = 5.983 xx 10^(24) kg Radius of the earth = 6.378 xx 10^(6) m G = 6.67 xx 10^(-11) Nm^(2) kg^(-2)

Calculate the mass and mean density of the earth from the following data : {:(,"Gravitational constant",(G),=6.6xx10^(-11)Nm^(2)kg^(-2)),(,"Radius of the earth",(R ),=6.37xx10^(6)m),(,"Acceleration due to gravity",(g),=9.8ms^(-2)):}