Acceleration due to Gravity on Moon

The expression for acceleration due to gravity is

Where G is the universal gravitational constant, M is the mass of the celestial body which produces acceleration in a body and R is the radius of the celestial body.

The equation for g shows that the value of acceleration due to gravity depends on the mass and radius of the celestial body and hence will be different for different celestial bodies.

Let us now derive a relation between the acceleration due to gravity on moon (g_m) and acceleration due to gravity on Earth (g_e).

Where M_e and R_e are the mass and radius of the Earth respectively.

Where M_m and R_m are the mass and radius of the moon respectively.

Divide equation (1) by equation (2)

We know that mass of the Earth is 100 times that of the moon and its radius is four times that of the moon.

i.e.,

M_e = 100 M_m

R_e = 4 R_m

Which means that acceleration due to gravity on moon is 1/6^th that on the Earth.

Numericals

01. The Earth's gravitational force causes an acceleration of 5 m/s² in a 1 kg mass somewhere in space. How much will the acceleration of a 3 kg mass be at the same place?

Solution:

The acceleration produced in any body due to the gravitational pull of the Earth does not depend on the mass of the body. So the acceleration produced in the

3 kg mass will also be 5 m/s².

02. Calculate the height of a bridge if a stone dropped from it takes six seconds to touch the surface of water.

Solution:

Initial velocity (u) = 0

Time taken (t) = 6 seconds

Acceleration due to gravity (g) = 9.8 m/s²

We make use of second equation of motion

h = 0 + 4.9 x 36

h = 176.4 m

03. A stone projected vertically upward, takes 5 seconds to reach the highest point. What is the initial velocity of the stone?

Solution:

Time taken = 5 seconds

Final velocity (v) = 0 (at maximum height velocity will be equal to zero)

v = u + gt

0 = u - 9.8 x 5

0 = u - 49.0

u = 49 m/s

Initial velocity = 49 m/s.