Question

Answer the following questions.

a. What is the difference between mass and weight of an object. Will the mass and weight of an object on the earth be same as their values on Mars? Why?

b. What are (i) free fall, (ii) acceleration due to gravity (iii) escape velocity (iv) centripetal force ?

c. Write the three laws given by Kepler. How did they help Newton to arrive at the inverse square law of gravity?

d. A stone thrown vertically upwards with initial velocity u reaches a height ‘h’ before coming down. Show that the time taken to go up is same as the time taken to come down.

e. If the value of g suddenly becomes twice its value, it will become two times more difficult to pull a heavy object along the floor. Why?

Answer 1

a. Difference between mass and weight of an object is as follows:

S. No.	Mass	Weight
1.	Mass is the amount of matter contained in a body.	Weight is the force exerted on a body due to the gravitational pull of another body such as Earth, the sun and the moon.
2.	Mass is an intrinsic property of a body.	Weight is an extrinsic property of a body.
3.	Mass is the measure of inertia.	Weight is the measure of force.
4.	The mass of a body remains the same everywhere in the universe.	The weight of a body depends on the local acceleration due to gravity where it is placed.
5.	The mass of a body cannot be zero.	The weight of a body can be zero.
6.	The SI unit of mass is kilogram (kg).	Since weight is a force, its SI unit is newton (N).
7.	The mass of a body can be measured using a beam balance and a pan balance.	The weight of a body can be measured using a spring balance and a weighing machine.

The mass of an object on the Earth will be same as that on Mars but its weight on both the planets will be different. This is because the weight (W) of an object at a place depends on the acceleration due to gravity of that place i.e. $W=mg\mathrm{or}W\propto g$ and since the values of acceleration due to gravity on both the planets differ, thus the weight of the object will be different for both the planets.

b. (i) A body is said to be under free fall when no other force except the force of gravity is acting on it.
(ii) The acceleration with which an object moves towards the centre of Earth during its free fall is called acceleration due to gravity. It is denoted by the letter ‘g’. It is a constant for every object falling on Earth’s surface.
(iii) The minimum velocity required to project an object to escape from the Earth's gravitational pull is known as escape velocity. It is given as:
$v_e=\sqrt{2gR}$
(iv) The force required to keep an object under circular motion is known as centripetal force. This force always acts towards the centre of the circular path.

c. Three laws given by Kepler is as follows:
First Law: The orbits of the planets are in the shape of ellipse, having the Sun at one focus.
Second Law: The area swept over per hour by the radius joining the Sun and the planet is the same in all parts of the planet’s orbit.
Third Law: The squares of the periodic times of the planets are proportional to the cubes of their mean distances from the Sun.

Newton used Kepler’s third law of planetary motion to arrive at the inverse-square rule. He assumed that the orbits of the planets around the Sun are circular, and not elliptical, and so derived the inverse-square rule for gravitational force using the formula for centripetal force. This is given as:
F = mv²/ r ...(i) where, m is the mass of the particle, r is the radius of the circular path of the particle and v is the velocity of the particle. Newton used this formula to determine the force acting on a planet revolving around the Sun. Since the mass m of a planet is constant, equation (i) can be written as:
F ∝ v²/ r ...(ii)
Now, if the planet takes time T to complete one revolution around the Sun, then its velocity v is given as:
v = 2πr/ T ...(iii) where, r is the radius of the circular orbit of the planet
or, v ∝ r/ T ...(iv) [as the factor 2π is a constant]
On squaring both sides of this equation, we get:
v² ∝ r²/ T²...(v)
On multiplying and dividing the right-hand side of this relation by r, we get:

$v^2\propto \frac{r^3}{T^2}\times \frac{1}{r}$ ...(vi)
According to Kepler’s third law of planetary motion, the factor r³/ T² is a constant. Hence, equation (vi) becomes:
v² ∝ 1/ r...(vii)
On using equation (vii) in equation (ii), we get:
$F\propto \frac{1}{r^2}$
Hence, the gravitational force between the sun and a planet is inversely proportional to the square of the distance between them.

d. For vertical upward motion of the stone:
S = h
u = u
v = 0
a = -g
Let t be the time taken by the ball to reach height h. Thus, using second equation of motion, we have
$$\begin{gathered} -2gh=v^2-u^2 \\ \Rightarrow u=\sqrt{2gh} \\ \mathrm{Now},\mathrm{from}\mathrm{first}\mathrm{equation}\mathrm{of}\mathrm{motion},\mathrm{we}\mathrm{have} \\ v=u-gt \\ t=\frac{u}{g}=\sqrt{\frac{2h}{g}}.....\left(\mathrm{i}\right) \end{gathered}$$
For vertical downward motion of the stone:
S = h
u = 0
a = g
Let v' be the velocity of the ball with which it hits the ground.
Let t' be the time taken by the ball to reach the ground. Thus, using second equation of motion, we have
$$\begin{gathered} 2gh=v{\text{'}}^2 \\ \Rightarrow v\text{'}=\sqrt{2gh} \\ \mathrm{Now},\mathrm{from}\mathrm{first}\mathrm{equation}\mathrm{of}\mathrm{motion},\mathrm{we}\mathrm{have} \\ v\text{'}=u+gt\text{'} \\ t\text{'}=\frac{v\text{'}}{g}=\sqrt{\frac{2h}{g}}.....\left(\mathrm{ii}\right) \end{gathered}$$
Hence, from (i) and (ii), we observe that the time taken by the stone to go up is same as the time taken by it to come down.

e. Let the mass of the heavy object be m. Thus, the weight of the object or the pull of the floor on the object is
W = mg
Now, if g becomes twice, the weight of the object or the pull of the floor on the object also becomes twice i.e. W' = 2mg = 2W
Thus, because of doubling of the pull on the object due to the floor, it will become two times more difficult to pull it along the floor.