NCERT Solutions Class 9 Science Chapter 9 Gravitation

2.0NCERT Questions with Solutions for Class 9 Science Chapter 9 - Detailed Solutions

• State the universal law of gravitation. Solution Every body in this universe attracts every other body with a force, which is directly proportional to the product of their masses and inversely proportional to the square of distance between their centres.
• Write the formula to find the magnitude of the gravitational force between the Earth and an object on the surface of the Earth. Solution $\mathrm{Fg}=\frac{\mathrm{GMm}}{{\mathrm{R}}^2}$ Where G is gravitational constant, M is the mass of the earth, $m$ is the mass of the object, R is the distance between the centre of the object and the centre of the Earth » radius of Earth.
• What do you mean by free fall? Solution When an object moves with a constant acceleration, under the influence of force of gravitation of the earth only, the object is said to have free fall.
• What do you mean by acceleration due to gravity? Solution The acceleration produced in a body due to force of gravity is called acceleration due to gravity. It is denoted by g. The value of acceleration due to gravity is taken as $9.8{\mathrm{ms}}^{-2}$ at the sea level.
• What are the differences between the mass of an object and its weight? Solution

• Why is the weight of an object on the Moon $\frac{1}{6}$ th of its weight on the Earth? Solution Acceleration due to gravity ( ${\mathrm{g}}_{\mathrm{m}}$ ) on Moon is $\frac{1}{6}$ th of the acceleration due to gravity (ge) on Earth. i.e., ${\mathrm{gm}}_{\mathrm{m}}=\frac{1}{6}{\mathrm{g}}_{\mathrm{e}}$ or ${\mathrm{mg}}_{\mathrm{m}}=\frac{1}{6}{\mathrm{mg}}_{\mathrm{e}}$ where, $\mathrm{m}=$ mass of object or ${\mathrm{W}}_{\mathrm{m}}=\frac{1}{6}{\mathrm{W}}_{\mathrm{e}}[$ weight $=\mathrm{mg}]$ Since, $g_m$ is 6th of $g_e$, thus, weight of an object on the Moon is also 6th of the weight of the object on Earth.
• You find your mass to be 42 kg on a weighing machine. Is your mass more or less than 42 kg ? Solution Mass is always a constant quantity. Therefore, it cannot be more or less than 42 kg .
• How does the force of gravitation between two objects change when the distance between them is reduced to half? Solution When all other variables remain constant, the force of gravitation is inversely proportional to the square of distance between the two objects. $\mathrm{F}\propto \frac{1}{{\mathrm{r}}^2}\Rightarrow \frac{{\mathrm{F}}^{\prime }}{\mathrm{F}}=\frac{{\mathrm{r}}^2}{{\left(\frac{\mathrm{r}}{2}\right)}^2}=\frac{4{\mathrm{r}}^2}{{\mathrm{r}}^2}=4$ or $F^{\prime }=4\mathrm{F}$ $\therefore$ The force of gravitation increases 4 times.
• Gravitational force acts on all objects in proportion to their masses. Why then, a heavy object does not fall faster than a light object? Solution A freely falling object of any mass falls under the action of gravity given by $\mathrm{g}=\frac{\mathrm{GM}}{{\mathrm{r}}^2}$, where ' G ' is constant of gravitation, M is mass of Earth, r is the distance between the object and the centre of Earth. Thus, the acceleration due to gravity Is independent of the mass of the objects. $\therefore$ All objects fall with the same acceleration towards the earth.
• What is the magnitude of the gravitational force between the Earth and a 1 kg object on its surface? (Mass of the Earth is $6\times 10^{24}\mathrm{kg}$ and radius of the Earth is $6.4\times 10^6\mathrm{m}$.) Solution ${\mathrm{Fg}}_{\mathrm{g}}=\frac{\mathrm{G}\times \mathrm{M}\times \mathrm{m}}{{\mathrm{R}}^2}$ Where, $M$ = mass of the Earth ; m = mass of object; R = radius of Earth. ${\mathrm{F}}_{\mathrm{g}}=\frac{6.67\times 10^{-11}\times \left(6\times 10^{24}\right)\times (1)}{{\left(6.4\times 10^6\right)}^2}$ $=9.770\mathrm{N}\approx 9.8\mathrm{N}$
• The Earth and the Moon are attracted to each other by gravitational force. Is the force with which the Earth attracts the Moon greater, smaller or the same as the force with which the Moon attracts the Earth? Why? Solution The Earth attracts the Moon with the same force with which the Moon attracts the Earth because, the gravitational forces between any two bodies are equal and opposite (Newton's third law).
• If the Moon attracts the Earth, why does the Earth not move towards the Moon? Solution The Earth does not move towards the Moon because the force exerted by the Earth or the Moon on each other is insufficient to move the Earth, on account of its huge mass.
• What happens to the force between two objects, if (i) the mass of one object is doubled? (ii) the distance between the objects is doubled and tripled? (iii)the masses of both objects are doubled? Solution (i) The force of gravitation doubles. (ii) The force of gravitation decreases 4 times if the distance between the objects is doubled, and if the distance between the objects is tripled then the force of gravitation decreases 9 times. (iii) The force of gravitation increases 4 times.
• What is the importance of universal law of gravitation? Solution Importance of universal law of gravitation is as follows : (i) It is the gravitational force between the Sun and the Earth, which makes the Earth move around the Sun. (ii) The tides formed in sea are because of gravitational pull exerted by the Sun and the Moon on the surface of water. (iii) It is the gravitational pull of Earth, which keeps us and other bodies firmly on the ground. (iv) It is the gravitational pull of the Earth, which holds our atmosphere in place.
• What is the acceleration of free fall? Solution The acceleration of free fall for object moving near the surface of Earth is $9.81{\mathrm{ms}}^{-2}$.
• What do we call the gravitational force between the Earth and an object? Solution It is called force of gravity.
• Amit buys few grams of gold at the poles as per the instruction of one of his friends. He hands over the same when he meets him at the equator. Will the friend agree with the weight of gold bought? If not, why? [Hint : The value of $g$ is greater at the poles than at the equator] Solution Weight of an object = mg, where ' m ' is mass of the object. ' $g$ ' at equator is less than the ' $g$ ' at poles. Thus, weight at equator will be less than that on poles. So, his friend will not agree with weight of the gold at the poles when measured at equator.
• Why will a sheet of paper fall slower than one that is crumpled into a ball? Solution Sheet crumpled into a ball has small surface area as compared to the similar unfolded sheet. Therefore, unfolded sheet will experience more resistance due to air as compared to the sheet crumpled into a ball, inspite of same force of gravity acting upon them. It is larger resistance of air which slows down the unfolded sheet, and therefore it falls slower as compared to sheet crumpled into a ball.
• Gravitational force on the surface of the Moon is only as strong as gravitational force on the Earth. What is the weight in newton of a 10 kg object on the Moon and on the Earth? Solution Mass of object (m) = 10 kg Acceleration due to gravity on Earth ( ${\mathrm{g}}_{\mathrm{e}}$ ) $=9.81{\mathrm{ms}}^{-2}$. Acceleration due to gravity on Moon ( ${\mathrm{g}}_{\mathrm{m}}$ ) $=\frac{9.81}{6}{\mathrm{ms}}^{-2}$. Weight of the object on the Earth $=mge=10\times 9.81=98.1\mathrm{N}$ Weight of the object on the Moon $={\mathrm{mg}}_{\mathrm{m}}=10\times \frac{9.81}{6}=16.35\mathrm{N}$
• A ball is thrown vertically upwards with a velocity of $49{\mathrm{ms}}^{-1}$. Calculate (i) The maximum height to which it rises. (ii) The total time it takes to return to the surface of the Earth. Solution (i) Initial velocity of the ball, $\mathrm{u}=9{\mathrm{ms}}^{-1}$ Final velocity of the ball, $\mathrm{v}=0$ Acceleration due to gravity $\mathrm{g}=-9.8{\mathrm{ms}}^{-2}$ [In upward direction, $g$ is taken -ve] Height attained by the ball, $\mathrm{s}=$ ? Time for rising up, $\mathrm{t}=$ ? We know, ${\mathrm{v}}^2-{\mathrm{u}}^2=2\mathrm{gs}$ $(0{)}^2-(49{)}^2=2\times (-9.8)\times s$ $\mathrm{s}=\frac{-49\times 49}{-2\times 9.8}=122.5m$ Alter : Maximum height achieved is given by, $H=\frac{u^2}{2g}=\frac{(49{)}^2}{2\times 9.8}=\frac{49\times 49}{2\times 9.8}=122.5\mathrm{m}$ (ii) We know $\mathrm{v}=\mathrm{u}+\mathrm{gt}$ $0=49-9.8\times t$ $\Rightarrow \mathrm{t}=\frac{49}{9.8}$ $\Rightarrow \mathrm{t}=5\mathrm{s}$ Now, time for upward journey of the ball $=$ the time for downward journey of the ball. $\therefore$ Total time taken by the ball to return to the surface of earth $=2\times t$ $=2\times 5=10\mathbf{s}$ [Aliter: Total time of journey, $\mathrm{T}=\frac{2\mathrm{u}}{\mathrm{g}}=\frac{2\times 49}{9.8}=10\mathrm{s}]$
• A stone is released from the top of a tower of height 19.6 m . Calculate its final velocity just before touching the ground. Solution Initial velocity of stone, $u=0$; final velocity of stone, $\mathrm{v}=$ ? height, $\mathrm{h}=19.6\mathrm{m}$; acceleration due to gravity, $\mathrm{g}=+9.8{\mathrm{ms}}^{-2}$ We know, ${\mathrm{v}}^2={\mathrm{u}}^2+2\mathrm{gh}$ ${\mathrm{v}}^2=(0{)}^2+2\times 9.8\times 19.6$ or ${\mathrm{v}}^2=19.6\times 19.6$ $\mathrm{v}=\sqrt{19.6\times 19.6}=19.6{\mathbf{ms}}^{-1}$.
• A stone is thrown vertically upward with an initial velocity of $40{\mathrm{ms}}^{-1}$. Taking $g=10{\mathrm{ms}}^{-}$ ${}^2$, find the maximum height reached by the stone. What is the net displacement and the total distance covered by the stone? Solution Initial velocity of stone $\mathrm{u}=40{\mathrm{ms}}^{-1}$; final velocity of stone $\mathrm{v}=0$; acceleration due to gravity $\mathrm{g}=-10{\mathrm{ms}}^{-2}$; [For upward direction $g$ is taken -ve]; height attained by stone ( h ) =? We, know, ${\mathrm{v}}^2-{\mathrm{u}}^2=2\mathrm{gh}$ or $(0{)}^2-(40{)}^2=2$ $\times (-10)\times h$ or $\mathrm{h}=\frac{-1600}{-20}=80\mathbf{m}$ $\therefore$ Maximum height attained by stone $=$ 80 m Net displacement of stone $=0$ (because the stone returns to the same point) Total distance covered by the stone $=2\times$ height attained $=2\times 80=160\mathbf{m}$
• Calculate the force of gravitation between the Earth and the Sun, given that the mass of the Earth $=6\times 10^{24}\mathrm{kg}$ and of the Sun $=2\times 10^{30}\mathrm{kg}$. The average distance between the wo is $1.5\times 10^{11}\mathrm{m}$. Solution Given mass of Earth, ${\mathrm{M}}_{\mathrm{e}}=6\times 10^{24}\mathrm{kg}$; mass of Sun, ${\mathrm{M}}_{\mathrm{s}}=2\times 10^{30}\mathrm{kg}$; distance between them, $\mathrm{r}=1.5\times 10^{11}\mathrm{m}$. Now, force of gravitation between them, $\mathrm{Fg}=\frac{{\mathrm{GM}}_{\mathrm{e}}{\mathrm{M}}_{\mathrm{s}}}{{\mathrm{r}}^2}$ $=\frac{6.67\times 10^{-11}\times 6\times 10^{24}\times 2\times 10^{30}}{1.5\times 10^{11}\times 1.5\times 10^{11}}$ $=35.57\times {10}^{21}\mathrm{N}$
• A ball thrown up vertically returns to the thrower after 6 s. Find. (a) the velocity with which it was thrown up (b) the maximum height it reaches, and (c) its position after $4\mathrm{s}($ take $\mathrm{g}=10\mathrm{m}/{\mathrm{s}}^2)$ Solution (a) The ball returns to the thrower in 6 s , thus, the time for its upward journey $=6$ $÷2=3\mathrm{s}$ For the upward motion of ball, initial velocity $u=$ ?; final velocity $\mathrm{v}=0$ ( Q Ball comes to rest) Time, $\mathrm{t}=3\mathrm{s}$ Acceleration due to gravity, $\mathrm{g}=-10{\mathrm{ms}}^{-2}$ [In upward direction, g is taken -ve] $\mathrm{v}=\mathrm{u}+\mathrm{gt}$ or $0=u-10\times 3$ or $-\mathrm{u}=-30$ or $u=30{\mathrm{ms}}^{-1}$ $\therefore$ Initial velocity of ball is $30{\mathbf{ms}}^{-1}$ (b) $s=ut+\frac{1}{2}g{t}^2$ $s=30\times 3-\frac{1}{2}\times 10\times (3{)}^2$ or $\mathrm{s}=90-45=45\mathrm{m}$ $\therefore$ Maximum height reached by ball is 45 m . (c) For the downward motion of ball, initial velocity, $u=0$; time for downward fall, $\mathrm{t}=4-3=1\mathrm{s}$; acceleration due to gravity, $\mathrm{g}=10{\mathrm{ms}}^{-2}$; distance covered in downward direction, $\mathrm{s}=$ ? $\mathrm{s}=\mathrm{ut}+\frac{1}{2}{\mathrm{gt}}^2$ $s=0\times (1)+\frac{1}{2}\times 10\times (1{)}^2$ $\Rightarrow \mathrm{s}=0+5$ $\Rightarrow \mathrm{s}=5\mathrm{m}$ $\therefore$ Position of ball after 4 s from ground $=45-5=40\mathrm{m}$.