Find the change in length of a second 's pendulum , if the acceleration due to gravity at the place changes from `9.75 m/s^2` to `9.8 m/s^2`.

Determine the length of a seconds pendulum at a place, where the acceleration due to gravity is 9.77 frac(m)(s^2)

The length of seconds pendulum at a place wl}ere g = 4.9 m/ s^2 is

Solve the following examples / numerical problems: The mass of a planet is 3 times the mass of the earth. Its diameter is 25600 km and the earth's diameter is 12800km. Find the acceleration due to gravity at the surface of the planet. [g (earth)=9.8 m/s^2]

If the density of the earth is tripled keeping its radius constant, then acceleration due to gravity will be (g=9.8 m/s^2)

An artificial satellite makes two revolutions per day around the earth. If the acceleration due to gravity on the surface of the earth is 9.8 m//s^2 and the radius of the earth is 6400 km, calculate the distance of satellite from the surface of the earth

Solve the following examples / numerical problems: If the acceleration due to gravity on the surface of the earth is 9.8 m/s^2, what will be the acceleration due to gravity on the surface of a planet whose mass and radius both are two times the corresponding quantities for the earth ?

If the density of the earth is doubled keeping its radius constant then acceleration due to gravity will be (g=9.8 m//s^(2))

Find the order of magnitude of the following quantities: Acceleration due to gravity=9.80665 m//s^2

(a) Assuming the earth to be a sphere of uniform density, calculate the value of acceleration due to gravity at a point, 1600 km above the earth, (b) Also find the rate of variation of acceleration due to gravity above the earth's surface. Radius of earth =6400 km, g =9.8 m//s^(2) .

Answer the following questions : What is the acceleration due to gravity at a height h (=radius of the earth) from the surface of the earth ? (g=9.8m/s^2)