The g is constant near the surface of earth. Therefore, all the equations for uniformly accelerated motion of bodies become valid when acceleration a is replaced by g.

**Modified equations** of motion are:

v = u + gt

s = ut + ½gt^{2}

v^{2} = u^{2} + 2gs

where u and v are the initial and final velocities and s is the distance covered in time t.

**Example 1:** Take the mass of the earth to be 6 ×10^{24} kg and its radius as 6.4 × 10^{6} m. Calculate the value of g. (G = 6.7 × 10^{–11} N m^{2} kg^{–2}).

g = GM/R^{2}

= 9.8 ms^{–2}

**Example 2:** The mass of the earth is 6 ×10^{24} kg and that of the moon is 7.4 × 10^{22} kg. If the distance between the earth and the moon is 3.84 × 10^{8} m, calculate the force exerted by the earth on the moon. G = 6.7 × 10^{–11} N m^{2} kg^{–2}.

The mass of the earth, m_{1} = 6 × 10^{24} kg

The mass of the moon, m_{2} = 7.4 ×10^{22} kg

The distance between the earth and the moon, r = 3.84 × 10^{8} m

G = 6.7 × 10^{–11} N m^{2} kg^{–2}

The force exerted by the earth on the moon is

F = Gm_{1}m_{2}/r^{2}

F = 2.01 × 10^{20} N

**Example 3:** A ball is thrown vertically upwards and rises to a height of 122.5 m. Calculate

- the velocity with which the ball was thrown upwards
- the time taken by the ball to reach the highest point

Take g = 9.8 ms^{–2}

Distance travelled, s = 122.5 m

Final velocity, v = 0 ms^{–1}

Acceleration due to gravity, g = 9.8 ms^{–2}

(i) From equation: v^{2} = u^{2} + 2gs

0 = u^{2} + 2(–9.8 ms^{–2}) × 122.5 m

For upward motion g is taken as negative.

–u^{2} = –2 × 9.8 × 122.5 m^{2}s^{–2}

u^{2} = 2401 m^{2}s^{–2}

u = 49 ms^{–1}

Thus, the velocity with which the ball was thrown upwards is 49 ms^{–1}.

(ii) From equation: v = u + gt

0 = 49 ms^{–1} + (9.8 ms^{–2}) × t

Therefore, t = 49/9.8 = 5s