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 + ½gt2
v2 = u2 + 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 ×1024 kg and its radius as 6.4 × 106 m. Calculate the value of g. (G = 6.7 × 10–11 N m2 kg–2).
g = GM/R2
= 9.8 ms–2
Example 2: The mass of the earth is 6 ×1024 kg and that of the moon is 7.4 × 1022 kg. If the distance between the earth and the moon is 3.84 × 108 m, calculate the force exerted by the earth on the moon. G = 6.7 × 10–11 N m2 kg–2.
The mass of the earth, m1 = 6 × 1024 kg
The mass of the moon, m2 = 7.4 ×1022 kg
The distance between the earth and the moon, r = 3.84 × 108 m
G = 6.7 × 10–11 N m2 kg–2
The force exerted by the earth on the moon is
F = Gm1m2/r2
F = 2.01 × 1020 N
Example 3: A ball is thrown vertically upwards and rises to a height of 122.5 m. Calculate
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: v2 = u2 + 2gs
0 = u2 + 2(–9.8 ms–2) × 122.5 m
For upward motion g is taken as negative.
–u2 = –2 × 9.8 × 122.5 m2s–2
u2 = 2401 m2s–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