Science Part i Solutions Solutions for Class 10 Science Chapter 1 - Gravitation

Answer:

Answer:

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

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.

Answer:

At the centre of Earth, the force due to upper half of the Earth will cancel the force due to lower half. In the similar manner, force due to any portion of the Earth at the centre will be cancelled due to the portion opposite to it. Thus, the gravitational force at the centre on any body will be 0. Since, from Newton's law, we know
F = mg
Since, mass m of an object can never be 0. Therefore, when F = 0, g has to be 0. Thus, the value of g is zero at the centre of Earth.

Answer:

From Kepler's third law of planetary motion, we have
$T^2\propto r^3$ .....(i)
Thus, when the period of revolution of planet at a distance R from a star is T, then from (i), we have
$T^2\propto R^3$ .....(ii)
Now, when the distance of the planet from the star is 2R, then its period of revolution becomes
$$\begin{gathered} T_1^2\propto (2R{)}^3 \\ \mathrm{or} \\ {T}_1^2\propto 8R^3 .....(\mathrm{iii}) \end{gathered}$$
Dividing (iii) by (ii), we get
$$\begin{gathered} \frac{T_1^2}{T^2}=\frac{8R^3 }{R^3} \\ \Rightarrow T_1=\sqrt{8}T \end{gathered}$$

Answer:

a. Here, u = 0 
S = 5 m
t = 5 s
From second equation of motion, we have
$$\begin{gathered} S=ut +\frac{1}{2}g{t}^2 \\ 5 =\frac{1}{2}g(5{)}^2 \\ g=\frac{10}{25}=0.4 \mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$
Hence, the value of g on the planet is 0.4 m/s².

b. The acceleration due to gravity of a planet is given as
$g=\frac{GM}{r^2}$
For planet A: $g_{\mathrm{A}}=\frac{G{M}_{\mathrm{A}}}{r_{\mathrm{A}}^2}$
For planet B: $g_{\mathrm{B}}=\frac{G{M}_{\mathrm{B}}}{r_{\mathrm{B}}^2}$
Now,
 $$\begin{gathered} g_{\mathrm{B}}=\frac{1}{2}{g}_{\mathrm{A}} \left(\mathrm{Given}\right) \\ \mathrm{or}, \frac{G{M}_{\mathrm{B}}}{r_{\mathrm{B}}^2}=\frac{G{M}_{\mathrm{A}}}{2r_{\mathrm{A}}^2} \\ \Rightarrow M_{\mathrm{B}}=\frac{M_{\mathrm{A}}{r}_{\mathrm{B}}^2}{2r_{\mathrm{A}}^2} \end{gathered}$$
Given: $r_{\mathrm{A}}=\frac{1}{2}{r}_{\mathrm{B}}$
$\Rightarrow M_{\mathrm{B}}=\frac{M_{\mathrm{A}}{r}_{\mathrm{B}}^2}{2({\displaystyle \frac{1}{2}}{r}_{\mathrm{B}}{)}^2}=2M_{\mathrm{A}}$
Thus, the mass of planet B should be twice that of planet A.

c. Mass of the object on Earth, m = 5 kg
Weight of the object on Earth, W_E = 49 N
Weight of the object on Moon,
$$\begin{gathered} W_{\mathrm{M}}=\frac{1}{6}{W}_{\mathrm{E}} \\ \Rightarrow W_{\mathrm{M}}=\frac{49}{6}=8.17 \mathrm{N} \end{gathered}$$ 
Mass of the object on Moon = 5 kg  (since mass is independent of the place of observation)

d. For vertical upward motion of the object,
S = 500 m
g = $-$10 m/s²
v = 0
Let u be the initial velocity of the object. From third equation of motion, we have
$$\begin{gathered} v^2-u^2=2aS \\ \Rightarrow 0-u^2=-2\times 10\times 500 \\ \Rightarrow u=100 \mathrm{m}/\mathrm{s} \end{gathered}$$
Now, let t₁ be time taken by the object to reach at 500 m height. Thus,
$$\begin{gathered} v=u+at \\ \Rightarrow 0=100-10\times t_1 \\ {t}_1=10 \mathrm{s} \end{gathered}$$
For vertical downward motion of the object,
S = 500 m
g = 10 m/s²
u = 0
Let t₂ be the time taken by the object to come back to the Earth from height of 500 m.
From second equation of motion, we have
$$\begin{gathered} S=ut+\frac{1}{2}a{t}^2 \\ \Rightarrow 500=\frac{10}{2}\times t_2^2 \\ \Rightarrow t_2=10 \mathrm{s} \end{gathered}$$
Thus, the total time taken by the object to reach back to Earth = t₁ + t₂ = 20 s

e. Here, t =1 s
g = 10 m/s²
u = 0
Let v be the velocity of the ball on reaching the ground.
Thus, from first equation of motion, we have
v = u + gt
$\Rightarrow$v = 10$\times$1 = 10 m/s
Hence, the speed of the object on reaching the ground is 10 m/s.
Let h be the height of the table. Thus, from second equation of motion, we have
$$\begin{gathered} S=ut+\frac{1}{2}g{t}^2 \\ \Rightarrow h=0+\frac{1}{2}\times 10\times 1^2 \\ \Rightarrow h=5 \mathrm{m} \end{gathered}$$
Hence, the height of the table is 5 m.

f. The gravitational force between the Moon and the Earth can be found out using the formula,
$F=G\frac{M_{\mathrm{e}}{M}_{\mathrm{m}}}{R^2}$
where, M_e and M_m are the masses of the Earth and the Moon, respectively. Using all the given values, we have
$F=(6.7\times {10}^{-11})\frac{(6\times {10}^{24})(7.4\times {10}^{22})}{(3.8\times {10}^5\times 1000{)}^2}=20.6\times {10}^{19}\simeq 2\times {10}^{20} \mathrm{N}$

g. The gravitational force between the Sun and the Earth can be found out using the formula,
$F=G\frac{M_{\mathrm{e}}{M}_{\mathrm{s}}}{R^2}$
where, M_e and M_s are the masses of the Earth and the Sun, respectively. Using all the given values, we have
$$\begin{gathered} 3.5\times {10}^{22}=(6.7\times {10}^{-11})\frac{(6\times {10}^{24})\times M_{\mathrm{s}}}{(1.5\times {10}^{11}{)}^2} \\ \Rightarrow M_{\mathrm{s}}=\frac{(3.5\times {10}^{22})\times (1.5\times {10}^{11}{)}^2}{(6.7\times {10}^{-11})\times (6\times {10}^{24})}=1.96\times {10}^{30 } \mathrm{kg} \end{gathered}$$