Question

Derive the relation between acceleration due to gravity (g) at the surface of a planet and gravitational constant (G).

Answer 1

From Newton's law of Gravitation, force is directly proportional to product of masses and inversely proportional to square of separation between masses.
e.g., F = $\frac{GmM}{r^2}$
Where m is mass of a object, M is mass of the earth and r is the separation between object and centre of the earth.

but from Newton's 2nd law of motion, force acts on object in gravitational field is given by,
F = mg [where g is acceleration due to gravity ]

Hence, mg = GmM/R²

Or, g = GM/R²

Hence, relation between g and G is given by $g=\frac{GM}{R^2}$

Answer 2

Answer:

Acceleration due to gravity in depth d from the earth's surface is given by, g=g_0\left(1-\frac{d}{R}\right)g=g

0

(1−

R

d

)

where g_0g

0

is the acceleration due to gravity on the earth's surface and R is the radius of the earth.

We have to find acceleration due to gravity at the centre of the earth.

at centre of the earth, depth = radius of earth's surface . e.g., d = R

Hence, g = g_0\left(1-\frac{R}{R}\right)g

0

(1−

R

R

)

g = 0

Hence, value of g will be zero at the centre of the earth.