Chapter - 10 Gravitation

The universal law of gravitation states that every object in the universe attracts every other object with a force called the gravitational force. The force acting between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers.

For two objects of masses m₁ and m₂ and the distance between them r, the force (F) of attraction acting between them is given by the universal law of gravitation as:

F = $\frac{G{m}_1{m}_2}{r^2}$

Where, G is the universal gravitation constant and its value is 6.67 × 10−11Nm2kg−2

Let M_E be the mass of the Earth and m be the mass of an object on its surface. If R is the radius of the Earth, then according to the universal law of gravitation, the gravitational force (F) acting between the Earth and the object is given by the relation:

F =  $\frac{G{M}_{E}m}{R^2}$

Gravity of the Earth attracts every object towards its centre. When an object is released from a height, it falls towards the surface of the Earth under the influence of gravitational force. The motion of the object is said to have free fall.

When an object falls towards the ground from a height, then its velocity changes during the fall. This changing velocity produces acceleration in the object. This acceleration is known as acceleration due to gravity (g). Its value is given by 9.8 m/s².

Let M_E be the mass of the Earth and m be an object on the surface of the Earth. Let R^E be the radius of the Earth. According to the universal law of gravitation, weight W_E of the object on the surface of the Earth is given by,

W_E = $\frac{G{M}_{E}m}{{R_E}^2}$

Let M_M and R_M be the mass and radius of the moon. Then, according to the universal law of gravitation, weight W_M of the object on the surface of the moon is given by:

W_M = $\frac{G{M}_{M}m}{{R_M}^2}$
Now,

⇒ $\frac{W_M}{W_E} = \frac{M_M {R_E}^2}{M_{E }{R_M}^2}$

Where,
ME = 5.98 × 1024 kg
MM = 7.36 × 1022 kg
RE = 6.4 × 106 m
RM = 1.74 × 106 m

⇒  $\frac{W_M}{W_E} =\frac{7.36 \times {10}^{22} \times {\left(6.4 \times {10}^6\right)}^2}{5.98 \times {10}^{24} \times {\left(1.74 \times {10}^6\right)}^2} = 0.165 = \frac{1}{6}$

Therefore, weight of an object on the moon is 1/6 of its weight on the Earth.

It is difficult to hold a school bag having a thin strap because the pressure on the shoulders is quite large. This is because the pressure is inversely proportional to the surface area on which the force acts. The smaller is the surface area; the larger will be the pressure on the surface. In the case of a thin strap, the contact surface area is very small. Hence, the pressure exerted on the shoulder is very large.

The upward force exerted by a liquid on an object immersed in it is known as buoyancy. When you try to immerse an object in water, then you can feel an upward force exerted on the object, which increases as you push the object deeper into water.

If the density of an object is more than the density of the liquid, then it sinks in the liquid. This is because the buoyant force acting on the object is less than the force of gravity. On the other hand, if the density of the object is less than the density of the liquid, then it floats on the surface of the liquid. This is because the buoyant force acting on the object is greater than the force of gravity.

When you weigh your body, an upward force acts on it. This upward force is the buoyant force. As a result, the body gets pushed slightly upwards, causing the weighing machine to show a reading less than the actual value.

The bag of cotton is heavier than iron bar. This is because the surface area of the cotton bag is larger than the iron bar. Hence, more buoyant force acts on the bag than that on an iron bar. This makes the cotton bag lighter than its actual value. For this reason, the iron bar and the bag of cotton show the same mass on the weighing machine, but actually the mass of the cotton bag is more than that of the iron bar.

According to the universal law of gravitation, gravitational force (F) acting between two objects is inversely proportional to the square of the distance (r) between them, i.e.,

F $\propto \frac{1}{r^2}$
If distance r becomes r/2, then the gravitational force will be proportional to

$\frac{1}{{\left({\displaystyle \frac{r}{2}}\right)}^2}$=$\frac{4}{r^2}$

Hence, if the distance is reduced to half, then the gravitational force becomes four times larger than the previous value.

All objects fall on ground with constant acceleration, called acceleration due to gravity (in the absence of air resistances). It is constant and does not depend upon the mass of an object. Hence, heavy objects do not fall faster than light objects.

According to the universal law of gravitation, gravitational force exerted on an object of mass m is given by

F = $\frac{GMm}{r^2}$

Where,
Mass of Earth, M = 6 × 1024 kg
Mass of object, m = 1 kg
Universal gravitational constant, G = 6.7 × 10⁻¹¹ Nm² kg⁻²
Since the object is on the surface of the Earth,
r = radius of the Earth (R)
r = R = 6.4 × 106 m
Therefore, the gravitational force

F =  $\frac{GMm}{r^2} = \frac{6.7 \times {10}^{-11}\times 6 \times {10}^{24} \times 1 }{{\left(6.4 \times {10}^6\right)}^2}$

= 9.8 N

According to the universal law of gravitation, two objects attract each other with equal force, but in opposite directions. The Earth attracts the moon with an equal force with which the moon attracts the earth.

The Earth and the moon experience equal gravitational forces from each other. However, the mass of the Earth is much larger than the mass of the moon. Hence, it accelerates at a rate lesser than the acceleration rate of the moon towards the Earth. For this reason, the Earth does not move towards the moon.

According to the universal law of gravitation, the force of gravitation between
two objects is given by F = $\frac{GMm}{r^2}$

(i) F is directly proportional to the masses of the objects. If the mass of one object is doubled, then the gravitational force will also get doubled.

(ii) F is inversely proportional to the square of the distances between the objects. If the distance is doubled, then the gravitational force becomes one-fourth of its original value.
Similarly, if the distance is tripled, then the gravitational force becomes one-ninth of its original value.

(iii) F is directly proportional to the product of masses of the objects. If the masses of both the objects are doubled, then the gravitational force becomes four times the original value.

The universal law of gravitation proves that every object in the universe attracts every other object.

When objects fall towards the Earth under the effect of gravitational force alone, then they are said to be in free fall. Acceleration of free fall is 9.8 ms⁻², which is constant for all objects (irrespective of their masses).

Gravitational force between the earth and an object is known as the weight of the object.

Weight of a body on the Earth is given by W = mg
Where,
m = Mass of the body
g = Acceleration due to gravity
The value of g is greater at poles than at the equator. Therefore, gold at the equator weighs less than at the poles. Hence, Amit’s friend will not agree with the weight of the gold bought.

When a sheet of paper is crumbled into a ball, then its density increases. Hence, resistance to its motion through the air decreases and it falls faster than the sheet of paper.

Weight of an object on the moon = $\frac{1}{6} \times$Weight of an object on the Earth

Also,
Weight = Mass × Acceleration
Acceleration due to gravity, g = 9.8 m/s²
Therefore, weight of a 10 kg object on the Earth = 10 × 9.8 = 98 N
And, weight of the same object on the moon = $\frac{1}{6} \times 98 = 16.3N$

(i) According to the equation of motion under gravity v² − u² = 2gs
Where,
u = Initial velocity of the ball
v = Final velocity of the ball
s = Height achieved by the ball
g = Acceleration due to gravity
At maximum height, final velocity of the ball is zero, i.e., v = 0 m/s and u = 49
m/s
During upward motion, g = − 9.8 m s⁻²
Let h be the maximum height attained by the ball.
Hence, using v² − u² = 2gs

We have, 0² − 49² = 2(−9.8)h ⇒ h = $\frac{49 \times 49}{2 \times 9.8}$ = 122.5 m

Let t be the time taken by the ball to reach the height 122.5 m, then according to
the equation of motion v = u + gt
We get,

0 = 49 + (−9.8)t ⇒ 9.8t = 49 ⇒ t = $\frac{49}{9.8} =5s$
But,
Time of ascent = Time of descent
Therefore, total time taken by the ball to return = 5 + 5 = 10 s

Answer 14:
According to the equation of motion under gravity v² − u² = 2gs

Where,
u = Initial velocity of the stone = 0 m/s
v = Final velocity of the stone
s = Height of the stone = 19.6 m
g = Acceleration due to gravity = 9.8 ms⁻²
∴ v² − 0² = 2 × 9.8 × 19.6
⇒ v² = 2 × 9.8 × 19.6 = (19.6)²
⇒ v = 19.6 ms⁻¹
Hence, the velocity of the stone just before touching the ground is 19.6 ms⁻¹

According to the equation of motion under gravity v² − u² = 2gs

Where,
u = Initial velocity of the stone = 40 m/s
v = Final velocity of the stone = 0 m/s
s = Height of the stone
g = Acceleration due to gravity = −10 m/s²
Let h be the maximum height attained by the stone.
Therefore, 0² − 40² = 2(−10)h ⇒ h = $\frac{40 \times 40}{20} = 80m$
Therefore, total distance covered by the stone during its upward and downward journey = 80 + 80 = 160 m
Net displacement during its upward and downward journey = 80 + (−80) = 0.

According to the universal law of gravitation, the force of attraction between the
Earth and the Sun is given by
F = $\frac{G \times M_{sun}\times M_{earth}}{R^2}$

Where,
M_sun = Mass of the Sun = 2 × 10³⁰ kg
M_earth = Mass of the Earth = 6 × 10²⁴ kg
R = Average distance between the Earth and the Sun = 1.5 × 10¹¹ m
G = Universal gravitational constant = 6.7 × 10⁻¹¹ Nm² kg⁻²

F = $\frac{6.7 \times {10}^{-11} 2 \times {10}^{30} \times 6 \times {10}^{24}}{{\left(1.5 \times {10}^{11}\right)}^2} = 3.57 \times {10}^{22}N$

Hence, the force of gravitation between the Earth and the Sun is 3.57 × 10²² N

Let the two stones meet after a time t.
When the stone dropped from the tower
Initial velocity, u = 0 m/s
Let the displacement of the stone in time t from the top of the tower be s.
Acceleration due to gravity, g = 9.8 m/s²
From the equation of motion,

$s = ut + \frac{1}{2}a{t}^2$

$s = 0 \times t + \frac{1}{2}\times 9.8\times t^2$

s= 4.9t²

When the stone thrown upwards
Initial velocity, u = 25 m/s
Let the displacement of the stone from the ground in time t be s′.
Acceleration due to gravity, g = −9.8 m/s²
Equation of motion,

$$\begin{gathered} s = ut + \frac{1}{2}a{t}^2 \\ s\text{'} = 25\times t - \frac{1}{2}\times 9.8 \times t^2 \end{gathered}$$

The combined displacement of both the stones at the meeting point is equal to the height of the tower 100 m.

s' + s =100

$\Rightarrow$25t - 4.9t² + 4.9t² =100

therefore, t = $\frac{100}{25}$s = 4 s

In 4 s, The falling stone has covered a distance given by (1) as s = 4.9 × 4² = 78.4 m
Therefore, the stones will meet after 4 s at a height (100 – 78.4) = 20.6 m from the ground.

(a) Time of ascent is equal to the time of descent. The ball takes a total of 6 s for its upward and downward journey.
Hence, it has taken 3 s to attain the maximum height.
Final velocity of the ball at the maximum height, v = 0 m/s
Acceleration due to gravity, g = −9.8 m/s²
Using equation of motion, v = u + at, we have
0 = u + (−9.8 × 3)
⇒ u = 9.8 × 3 = 29.4 m/s
Hence, the ball was thrown upwards with a velocity of 29.4 m/s.

(b) Let the maximum height attained by the ball be h.
Initial velocity during the upward journey, u = 29.4 m/s
Final velocity, v = 0 m/s
Acceleration due to gravity, g = −9.8 ms−2

Using the equation of motion, 

$$\begin{gathered} s = ut + \frac{1}{2}a{t}^2 \\ h = 29.4 \times 3 - \frac{1}{2}\times 9.8 \times 3^2 \end{gathered}$$

⇒ h = 44.1 m

Hence, the maximum height is 44.1 m.

(c) Ball attains the maximum height after 3 s. After attaining this height, it will start falling downwards.
In this case,
Initial velocity, u = 0 m/s
Position of the ball after 4 s of the throw is given by the distance travelled by it during its downward journey in 4 s − 3 s = 1 s.
Using the equation of motion,

$s = ut + \frac{1}{2}a{t}^2$

s = 0 × 1 + $\frac{1}{2}\times 9.8\times 1$² ⇒ s = 4.9 m

Now, total height = 44.1 m
This means, the ball is 39.2 m (44.1 m − 4.9 m) above the ground after 4 seconds.

An object immersed in a liquid experiences buoyant force in the upward direction.

Two forces act on an object immersed in water. One is the gravitational force, which pulls the object downwards, and the other is the buoyant force, which pushes the object upwards. If the upward buoyant force is greater than the downward gravitational force, then the object comes up to the surface of the water as soon as it is released within water. Due to this reason, a block of plastic released under water comes up to the surface of the water.

If the density of an object is more than the density of a liquid, then it sinks in the liquid. On the other hand, if the density of an object is less than the density of a liquid, then it floats on the surface of the liquid.
Here, density of the substance = $\frac{mass of the subs\tan ce}{volume of the subs\tan ce} = \frac{50}{20} = 2.5 g/c{m}^3$

The density of the substance is more than the density of water (1 g cm⁻³).

Hence, the substance will sink in water.

Density of the 500 g sealed packet = $\frac{Mass of the packet}{Volume ofthe packet} = \frac{500}{300} = 1.428 g/c{m}^3$

The density of the substance is more than the density of water (1 g/cm³). Hence, it will sink in water.
The mass of water displaced by the packet is equal to the volume of the packet, i.e., 350 g.