Page 114 - Understanding NCERT Science 09

change in the velocity of falling objects? While calculations, we can take g to be more or less
falling, there is no change in the direction of constant on or near the earth. But for objects
motion of the objects. But due to the earth’s far from the earth, the acceleration due to
attraction, there will be a change in the gravitational force of earth is given by
magnitude of the velocity. Any change in Eq. (9.7).
velocity involves acceleration. Whenever an
object falls towards the earth, an acceleration 9.2.1 TO CALCULATE THE VALUE OF g
is involved. This acceleration is due to the
earth’s gravitational force. Therefore, this To calculate the value of g, we should put the
acceleration is called the acceleration due to values of G, M and R in Eq. (9.9), namely,
the gravitational force of the earth (or universal gravitational constant, G = 6.7 × 10 –
24
-2
2
acceleration due to gravity). It is denoted by 11 N m kg , mass of the earth, M = 6 × 10 kg,
6
g. The unit of g is the same as that of and radius of the earth, R = 6.4 × 10 m.
–2
acceleration, that is, m s .
M
We know from the second law of motion g = G 2
that force is the product of mass and R
acceleration. Let the mass of the stone in -11 2 -2 24
´
´
6.7 10 N m kg ´ 6 10 kg
activity 9.2 be m. We already know that there =
6
(6.4 10 m) 2
´
is acceleration involved in falling objects due
to the gravitational force and is denoted by g. = 9.8 m s .
–2
Therefore the magnitude of the gravitational
force F will be equal to the product of mass Thus, the value of acceleration due to gravity
–2
of the earth, g = 9.8 m s .
and acceleration due to the gravitational
force, that is,
F = m g (9.6) 9.2.2 MOTION OF OBJECTS UNDER THE
From Eqs. (9.4) and (9.6) we have INFLUENCE OF GRAVITATIONAL
M ´m FORCE OF THE EARTH
m g = G 2
d Let us do an activity to understand whether
all objects hollow or solid, big or small, will
M fall from a height at the same rate.
or g = G 2 (9.7)
d
Activity ______________ 9.3
where M is the mass of the earth, and d is the
distance between the object and the earth.
• Take a sheet of paper and a stone.
Let an object be on or near the surface of
Drop them simultaneously from the
the earth. The distance d in Eq. (9.7) will be first floor of a building. Observe
equal to R, the radius of the earth. Thus, for whether both of them reach the
objects on or near the surface of the earth, ground simultaneously.
• We see that paper reaches the ground
M × m
mg = G 2 (9.8) little later than the stone. This happens
R because of air resistance. The air offers
resistance due to friction to the motion
M of the falling objects. The resistance
g = G 2 (9.9)
R offered by air to the paper is more than
the resistance offered to the stone. If
The earth is not a perfect sphere. As the we do the experiment in a glass jar
radius of the earth increases from the poles to from which air has been sucked out,
the equator, the value of g becomes greater at the paper and the stone would fall at
the poles than at the equator. For most the same rate.

GRAVITATION 103

Rationalised 2023-24