Question

a body weighs 63N on the surface of earth.what is GRAVITATIONAL force on it due to the height of 3200 km from earth surface. help me .​

Answer 1

Answer:

The initial force exerted is F = mg = 63 N

so the mass of the body would be m = F/g = 63/10 kg = 6.3 kg

Now, the variation in g due to height is

g1 = g / (1 + (h/R))^2

R is the radius of Earth

g is acceleration due to gravity close to Earth's surface (or on it)

here h is half the radius of the Earth or h = R/2

so,

g1= g(1 + R/2R)^2

g1 = g / (3/2)^2

or g1

= (4/9)g = 4.44 m/s^2

Now, the force experienced by the body of mass 6.3 kg at height half the radius of Earth would be

F1 = 6.3 x 4.4 = 27.72 N

Answer 2

Answer:

Weight of the body, W = 63 N

Acceleration due to gravity at height hfrom the Earth’s surface is given by the relation:

g‘ = g / [1 + ( h / Re) ]2

Where,

g = Acceleration due to gravity on the Earth’s surface

Re = Radius of the Earth

For h = Re / 2

g‘ = g / [(1 + (Re / 2Re) ]2

= g / [1 + (1/2) ]2 = (4/9)g

Weight of a body of mass m at height h is given as:

W‘ = mg

= m × (4/9)g = (4/9)mg

= (4/9)W

= (4/9) × 63 = 28 N.