Gravitation Class 9 Science | NCERT Science Chapter 10 | In-Text Questions & Answers | Exercises Questions & Answers | Notes & pdf
♦ Gravitation Science Chapter 10 – We learnt acceleration due to gravity in Q5 in Motion chapter
Q1. What is universal law of gravitation?
The universal law of gravitation states that the force of the attraction between two objects in the universe is proportional to the product of their masses and inversely proportional to the square of the distance between them. Let assume two objects X and Y of masses M and m respectively. The distance between them is d. The force of attraction between them is F.
Firstly, as per universal law of gravitation, F is proportional to the product of M and m –
F ∝ Mxm
Secondly, as per universal law of gravitation, F is inversely proportional to the square of the distance d –
F ∝ 1/d2
After combining the aforesaid two equations –
⇒ F ∝ (Mxm)/d2
⇒ F = G [(Mxm)/d2 ], where G is the constant of proportionality.
G called universal gravitation constant.
Q2. What is the accepted value of universal gravitation constant G?
Accepted value of the universal gravitation constant G is 6.673 x 10-11 Nm2 Kg-2. It is established by the English scientist Henry Cavendish.
Q3. What is the unit of universal gravitation constant G?
As per universal law of gravitation –
⇒ F = G(Mxm)/d2
⇒ Fxd2 = G(Mxm)
So, G = (Fxd2 )/(Mxm)
Therefore, unit of G = (Newton x metre2 )/(KgxKg) = Nm2/Kg2 = Nm2Kg-2.
Q4. Find the gravitational force between the Sun & the Earth. Given that ⇒ mass of the Sun = 1.989×1030 kg, mass of the earth = 6×1024 kg, distance between the earth and the sun =1476×108 m, universal gravitation constant G = 6.673×10-11 Nm2Kg-2.
Gravitational force between sun and earth :
⇒ Fsun-earth = G [(Msun x mearth)/(dsun-earth ) 2],
⇒ Fsun-earth = 6.673×10-11 [(1.989 x 1030 x 6 x 1024)/(1476 x 108) 2] N = 3.65×1022 N.
Q5. Find the gravitational force between the Earth and the Moon. Given that ⇒ mass of the Earth = 6×1024 kg, mass of the Moon = 7.4×1022 kg, distance between earth and moon = 3.84×108 m, universal gravitation constant G = 6.673×10-11 Nm2Kg-2.
Gravitational force between earth and moon :
⇒ Fearth-moon = G[( Mearth x mmoon)/(dearth-moon ) 2],
⇒ Fearth-moon = 6.673 x 10-11 [(6 x 1024 x 7.4 x 1022)/(3.84 x 108) 2] N = 20.09 x 1019 N.
Q6. Explain the gravitational force in brief.
Concept of gravitation is first conjectured by the great scientist Sir Isaac Newton which is related to famous apple story. An apple falls on Newton when he was sitting under an apple tree. That incident arises some questions in Newton’s mind. Newton thinks that by which attraction force apple falls on earth. If earth can attract an apple, does it attract everything around it? Does it attract moon? Which force is responsible in all these cases etc.? In our solar system every planets rotate around sun. Similarly secondary planet or satellite rotates around main planet like moon rotates around earth. Throwing of an object upwards or dropping of an object from a height always results returning of that object to earth. The attraction force responsible for all these actions is the gravitational force.
Q7. Explain the centripetal force in brief.
Centre-seeking force or centripetal force is the force which causes rotation of an object along fixed orbit around a centralised object. India, USA, Russia, China, France and some other countries launch satellites for various purposes. These satellites rotate around earth by moving along certain orbit due to centripetal force. Rotation of moon around earth is also an example of centripetal force. Basically, the gravitational force i.e. attraction force between the orbital object and the centralised object causes this centripetal force. Actually every object in our universe is attracting each other due to gravitational force.
Q8. Why does not an apple fall on earth when Newton’s third law of motion states that every action has an equal and opposite reaction?
An apple falls on earth due to gravitational force. If earth attracts apple, apple also attracts earth as Newton’s third law states that every action has an equal and opposite reaction. Then why apple falls on earth? Why not earth moves to apple? We know from Newton’s second law of motion –
F = ma, ⇒ a = acceleration = F/m = Force / Mass.
As per Newton’s third law, by the force earth attracts apple is equal to the force by which apple attracts earth.
So, acceleration of earth = Force / Mass of the earth …………….Case#1.
Similarly, acceleration of the apple = Force / Mass of the apple ……………….Case#2.
Now compare the mass of an apple and the mass of the earth. The mass of the earth (5.8 x 1024 kg) is so huge that mass of an apple (few grams) is negligible. As a result, the value obtained from Case#2 is definite but value obtained from Case#1 is almost zero. That’s why apple falls on earth. We may apply same principle to our planetary system and realise how everything is balanced due to gravitational force.
Q9. What is free fall?
Earth always attracts nearby objects due to gravitational force. That’s why fruit from tree falls on ground, even if you throw anything towards sky it returns back to earth. When any object falls from the upper place towards ground, we may term it as free falling under the influence of the gravitational force.
1. A person drops an object from an high-rise which touches earth after 5 s. Find the height of the high-rise. What will be speed of the object when it touches earth
Distance covered in 5 s by the object = height of the high-rise. Take ɡ= 9.8 ms-2. Also here initial velocity = u = 0.
So height of the high-rise = s = ut + ½ ɡt2 = 0x5 + ½ x 9.8 x 52 m = 122.5 m.
Speed of the object when it touches earth = v = u + ɡt = 0 + 9.8×5 ms-1 = 49 ms-1.
2. A person throws an object upwards and the object travels for 10 s. Find the height the object reaches. Find the initial velocity. Take ɡ= 10 ms-2 .
ɡ = 10 ms-2, t = 10 s, initial velocity u = ? Final velocity = v = 0.
Now, v = u – ɡt (for negative acceleration)
⇒ So initial velocity, u = v + ɡt = 0 + 10×10 m/s = 100 m/s.
Therefore, v2 = u2 – 2ɡs (∵ negative acceleration)
⇒ s = (u2 – v2)/2ɡ
⇒ height reached by the object = S = (u2 – v2)/ 2ɡ = (1002 – 02 )/2×10 m = 500 m.
Q10. What is acceleration due to gravity?
Suppose a mango is falling from a tree. When it tears apart itself from the tree, we find it’s initial velocity is nil. But when mango is falling on earth, it is falling with a velocity. Therefore, from the time of starting of falling to the time of touching of the earth, there are some changes in velocity. As a result, we say that any free falling involves some changes in velocity. Also we know that any change in velocity is associated with acceleration. So the velocity of falling object is actually accelerated velocity. This acceleration is created by the gravity i.e. gravitational force and is denoted by ɡ. The unit of ɡ is m/s2 or ms-2 i.e. same as that of acceleration. The value of ɡ is 9.8 ms-2.
Equation of g : Suppose an object of mass m is in the mode of free falling. Now gravitational force acting between earth and the object is ⇒ F = G( M x m )/d2.
(where M = mass of the earth, d = distance between earth and the object, m = mass of the object, G = universal gravitation constant).
Now acceleration associated with the object due to gravitational force is ɡ. We also know ⇒ Force = mass x acceleration. As acceleration ɡ is induced by the gravitational force, so product of [m x ɡ] is equal to the gravitational force i.e. G(Mxm)/d2. So
⇒ m x ɡ = G(Mxm)/d2,
⇒ ɡ = GM/d2 .
When the object is just touching the earth, d becomes equal to the radius of the earth. Let take R as the radius of the earth and we may write, ɡ = GM/d2 = GM/R2 .
Q11. Find the value of acceleration due to gravity.
From above discussion ⇒ ɡ = GM/R2 .
Putting the values of G = universal gravitation constant = 6.7 x 10-11 N m2 Kg-2, M = mass of the earth = 6 x 1024 Kg, R = radius of the earth = 6.4 x 106 ⇒
⇒ ɡ = GM/R2 = [ (6.7 x 10-11 x 6 x 1024) / (6.4 x 106 )2 ] ms-2 = 9.8 ms-2.
Value of ɡ is not same all over the earth as our earth is not perfect sphere and the radius of the earth (R) varies from polar area to equator line, mountain, mines etc.
Q12. Can we apply equations of motion with uniform acceleration in case of free falling objects?
We get following equations from Motion chapter –
A. s = ut + ½ at2
B. v2 = u2 + 2as
C. v = u + at
Now replace acceleration a by acceleration due to gravity ɡ . Then we get following equations–
A. s = ut + ½ ɡt2
B. v2 = u2 + 2ɡs
C. v = u + ɡt
All equations satisfy the motion of a free falling object.
Q13. What is mass?
From the discussion of Forces and Laws of Motion chapter, we get in conclusion that mass represents measurement of inertia of an object. The more the inertia the more is the mass. The C.G.S. unit of mass is gram and M.K.S. or S.I. unit is kilo gram. Mass of an object remains same and unaltered at anywhere in the universe be it on earth or moon or other planets.
Q14. What is weight and what is unit of weight?
Earth attracts every object towards it and weight of an object is the measurement of the attraction force by the earth towards the object. That attraction force (F) is equal to (m x ɡ) where ɡ is acceleration due to gravity. So weight of an object is ⇒
W = F = m x ɡ (m= mass of the object, ɡ = acceleration due to gravity).
Unit of weight is same as that of force and it is Newton.
Q15. Does weight change in earth?
The value of ɡ is not fixed all over the earth as our earth is not perfect sphere, so the radius of the earth varies from polar area to equator line to mountain to mines etc. As ɡ varies, so W varies (as mass m remains same at anywhere).
⇒ W = F = m x ɡ ,
⇒ W ∝ ɡ .
So weight is proportional to acceleration to gravity.
♦ e.g. Mass of an object on earth surface is 50 kg. Find it’s weight.
Weight = m x ɡ . Putting value of ɡ= 9.8 ms-2, we get-
Weight = m x ɡ = 50 x 9.8 N = 490 N.
Q16. What is the difference between N (Newton) and kgf?
Newton is the unit of force and 1 N force is the force which induces acceleration of 1 ms-2 in an object of 1 kg mass. So ⇒ 1 Newton = 1 kg x 1 ms-2
Again, weight = W = F = m x ɡ [m= mass of the object, ɡ= acceleration due to gravity].
After putting value of ɡ = 9.8 ms-2 , m = 1 kg, we get-
F = m x ɡ = 1 kg x 9.8 ms-2 = 9.8 N ≃ 10 N. So, 1 kgf ≃10N.
This 10 N value of F is known as 1 kgf. You will learn kgf unit in higher classes.
Q17. Compare the weight of an object in earth and the weight of that object in moon.
Basically, weight (W) of an object of mass m is the attraction force by the earth towards that object. W = F = mxɡ (m= mass of the object, ɡ = acceleration due to gravity). This force is basically the gravitational force between the object and the earth. So, W = G(M x m)/R2 (M= mass of the earth, R = radius of the earth, G= universal gravitation constant). We may write this equation as –
Wearth = GMearth x m / R2earth ———— Equation 1.
Similarly, weight (W) of an object of mass m in moon is the gravitational force between the object and the moon. And we may write Equation 1 for moon as follows ⇒
Wmoon = GMmoon x m / R2moon ————– Equation 2.
Comparison of weight of an object in earth and moon ⇒
After dividing Equation 1 by Equation 2 –
⇒ Wearth / Wmoon = [GMearth x m / R2earth ]/[GMmoon x m / R2moon]
⇒ Wearth / Wmoon = (Mearth x R2moon)/(Mmoon x R2earth).
After putting values ⇒
Mearth = mass of the earth = 5.98 x 1024 kg , Rearth = radius of the earth = 6370 km = 6370 x103 m,
Mmoon = mass of the moon = 7.36 x 1022 kg , Rmoon = radius of the moon = 1740 km = 1740 x 103 m.
The equation becomes ⇒
⇒ Wearth / Wmoon = [5.98 x 1024 x (1740 x 103)2] / [7.36 x 1022 x (6370 x 103)2 ]
⇒ Wearth / Wmoon = [18.105 x 1036 ]/[ 2.986 x 1036 ] = 6.06 ≃ 6.
In conclusion, weight of an object in earth = 6 x weight of an object in moon. Therefore, weight of any object in moon is 6 times lighter than in earth. In other words, weight of any object in earth is 6 times heavier than in moon.
Q18. Compare the weight of an object in earth and the weight of that object in mars.
Basically, weight (W) of an object of mass m is the attraction force by the earth towards that object. Therefore, W = F = m x ɡ (m= mass of the object, ɡ = acceleration due to gravity). This force is basically the gravitational force between the object and the earth. So, W = G(M x m)/R2 (M= mass of the earth, R = radius of the earth, G= universal gravitation constant). We may write this equation as –
Wearth = G Mearth x m / R2earth ———— Equation 1.
Similarly, weight (W) of an object of mass m in mars is the gravitational force between the object and the mars. We may write Equation 1 for moon as follows ⇒
Wmar = G Mmar x m / Rmar —————– Equation 3.
Comparison of weight of object in earth and mars ⇒
After dividing Equation 1 by Equation 3 –
⇒ Wearth / Wmar = [GMearth x m / R2earth ] / [G Mmar x m / R2mar ]
⇒ Wearth / Wmar = (Mearth x R2mar ) / (Mmar x R2earth).
After putting values ⇒
Mearth = mass of the earth = 5.98 x 1024 kg , Rearth = radius of the earth = 6370 km = 6370 x 103m ,
Mmar = mass of the mars = 6.39 x 1023 kg , Rmar = radius of the mars = 3389 km = 3389 x 103m.
The equation becomes-
⇒ Wearth / Wmar = [ 5.98 x 1024 x (3389 x 103)2 ] / [6.39 x 1023 x (6370 x 103)2 ]
⇒ Wearth / Wmar = [ 68.682 x 1036 ] / [ 25.928 x 1036 ] = 2.648 ≃ 2.65.
In conclusion, weight of an object in earth = 2.65 x weight of an object in mars. Therefore, weight of any object in mars is 2.65 times lighter than in earth. In other words, weight of any object in earth is 2.65 times heavier than in mars.
Q19. What is thrust? What is pressure?
Put a iron cube of mass 5 kg on a table. Also keep a block of wood of mass 5 kg beside the iron cube. Both the object are exerting force vertically downwards equal to their body weight i.e. 5 x g N. You see that surface area covered by the iron cube is much less than the block of wood but forces exerted by both the object are same i.e. 5 x g N.
Thrust is the force acting vertically downwards i.e. perpendicular to the surface and pressure is the thrust per unit area. N/m2 = Nm-2 is the S.I. unit of pressure. 1 N/m2 is also called 1 Pascal (named after scientist Blaise Pascal). Dyne/cm2 is the C.G.S. unit of pressure. Generally, solid exerts pressure on a surface where it is kept on and fluids i.e. liquids and gases exert pressure in all direction of it’s container.
Q20. What is buoyancy?
Buoyancy is the upthrust force of a fluid experienced by an object when the object is immersed in that fluid. Take a bucket of water. Drop a pebble in it and the pebble sinks into it easily. Now drop an empty plastic water bottle in the bucket and the bottle floats. Try to push the bottle into the water. You feel an upward force acting against your push. This upthrust or upward force is known as buoyant force.
The buoyant force increases more if you try to push the bottle deeper into the water. Then why the pebble sinks so easily? We know that weight of an object is the gravitational attraction force of the earth acting on the object. Sinking of an object depends on this gravitational attraction force of the earth acting on that object (i.e. weight of the object). If it is greater than the buoyant force of the fluid, then it sinks into the fluid. If not, then it floats.
That’s why a pebble easily sinks into water as it’s weight is more than the upthrust or upward force exerted by the water. In case of water bottle, the weight of water bottle (i.e. the gravitational attraction force of the earth acting on the water bottle) is less than the upthrust or upward force exerted by the water. So the bottle floats on the water. Now if you want to immerse fully the bottle into water, your applied force must be equal or more than the difference between the buoyant force and weight of the bottle. Density of the fluid determines the buoyant force or upthrust of that fluid.
Q21. What is density?
Density of any substance is the mass per unit volume of that substance under specified condition. S.I. unit of density is kg/m3. For examples, density of various matter – gold ⇒ 19300 kg/m3, copper ⇒ 8944 kg/m3, water ⇒ 1000 kg/m3, iron ⇒ 7860 kg/m3 etc. Density of gold is 19300 kg/m3 and it means that a cube of gold (i.e. 1 m3 gold) weighs 19300 kg.
Density = mass/unit volume & S.I. unit of density = kg/m3
Q22. How density is related to buoyancy?
If the density of an object is more than the density of the fluid, then the object will sink into the fluid. And If the density of an object is less than the density of the fluid, then the object will float on the fluid.
Q23. What is relative density?
Relative density of any substance is the ratio of it’s density to the density of water. As it is the ratio of two density, so it has no unit. We know density of gold is 19300 kg/m3 and density of water is 1000 kg/m3. So relative density of gold ⇒ (19300/1000) ⇒ 19.3.
Relative density of a substance = Density of that substance/Density of water
Relative density is a ratio of two density. So it has no unit.
Q24 . What is Archimedes’ Principle about buoyancy?
Famous ‘Eureka!’ word was the first reaction when the great Greek scientist Archimedes discovered the principle regarding buoyancy. He observed that water overflows when he steps in bathtub. As per Archimedes’ Principle ‘When an object or body is fully or partially immersed in a fluid, it displaces some fluid and experiences an upthrust or upward force which is equal to the weight of displaced fluid’. When you swim, you feel lighter yourself, why? Because when you step into water, your body displaces some water and that displaced water reduces your weight equal to the weight of the displaced water.
Take a bucket of water and a piece of stone. The piece of stone is fitted with spring balance which shows it’s weight. Let sink that into water. We see that reading in spring balance reduces and level of water increase. As soon as the piece of stone is put into water, it displaces some water equal to it’s volume which results increase in water level.
We also see that reading in spring balance reduces. This reduction in weight is due to upthrust or buoyant force given by the water on the piece of stone. And as per Archimedes’ Principle, this reduction in weight of the stone is equal to the weight of the water displaced by the stone.
In-Text Questions & Answers | NCERT Chapter 10 | Gravitation Class 9 Science
In-Text Questions & Answers | NCERT Chapter 10 | Gravitation Class 9 Science | Page Nos. 134
1. State the universal law of gravitation.
Write down the answer of Q1 in detail.
2. Write the formula to find the magnitude of the gravitational force between the earth & an object on the surface of the earth.
As per universal law of gravitation ⇒ F = G (Mxm)/d2, (where G = universal gravitation constant). Let assume, Mearth = mass of the earth, mobject = mass of an object on the surface of the earth, d = R = Radius of the earth = distance between the earth & the object on the surface of the earth. Then, the magnitude of gravitational force is ⇒ F = G (Mearth x mobject)/R2.
In-Text Questions & Answers | NCERT Chapter 10 | Gravitation Class 9 Science | Page Nos. 136
1. What do you mean by free fall?
Write down the answer of Q9 in detail.
2. What do you mean by acceleration due to gravity?
Write down the answer of Q10 in detail.
In-Text Questions & Answers | NCERT Chapter 10 | Gravitation Class 9 Science | Page Nos. 138
1. What are the differences between the mass of an object & its weight?
| Differences between mass & weight - | |
|---|---|
|
Mass |
Weight |
|
It is the quantity of matter that a body contains. |
It is the gravitational force acting on the body. |
|
Basically mass is the inertia of the body. |
Basically weight s the measure of the gravity. |
|
It is constant & does not vary during changing of place. |
It is not constant & varies when place is changed. |
|
Mass has only magnitude but no direction. So it is a scalar quantity. |
Weight has both magnitude & direction. So it is a vector quantity. |
|
Kilogram i.e. kg is the S.I. unit of mass. |
Newton (i.e.) N is the S.I. unit of weight. |
2. Why is the weight of an object on the moon 1/6 th its weight on the earth?
Write down Q17 in detail.
In-Text Questions & Answers | NCERT Chapter 10 | Gravitation Class 9 Science | Page Nos. 141
1. Why it is difficult to hold a school bag having a strap made of a thin & strong string?
The bag exerts force on tiny & strong string. That force exerted equally by the string exerts a very less area on our body. We know that pressure is inversely proportional to its area. Therefore, force exerted on our body by the thin string of the school bag will be more. So it is tough to carry the bag.
2. What do you mean by buoyancy?
Buoyancy is the upward force exerted by a liquid on an object immersed in it. When you try to immerse an object into water, you feel an upward or upthrust force exerted on the object. The more you try to push the object deeper into water, the more pressure you will feel.
3. Why does an object float or sink when placed on the surface of water?
When the density of an object is more than the density of water, then the object will sink in water. In this case, the buoyant force acting on the object is less than the force of gravity acting on it. On the other side, if the density of an object is less than the density of water, then the object will float on the water. In this case, the buoyant force acting on the object is more than the force of gravity acting on it.
In-Text Questions & Answers | NCERT Chapter 10 | Gravitation Class 9 Science | Page Nos. 142
1. You find your mass to be 42 kg on a weighing machine. Is your mass more or less than 42 kg?
It is indeed less than 42 kg. When you put your body on a weight machine, your body gets pushed slightly upwards. Because an upward force acts on it. Basically this force is the buoyant force. Therefore, the result showed by the weight machine is less than the actual value.
2. You have a bag of cotton & an iron bar, each indicating a mass of 100 kg when measured on a weighing machine. In reality, one is heavier than other. Can you say which one is heavier & why?
An iron bar is lighter than a cotton bag of same mass. Because the surface area of a cotton bag is more than an iron bar. So the buoyant force on the cotton bag will be more than that on the iron bar. Therefore, the cotton bag is found lighter than its actual value. When you weigh both the items, weighing machine shows same mass. But actually the mass of the cotton bag is more than that of the iron bar.
Exercises Questions & Answers | NCERT Chapter 10 | Gravitation Class 9 Science | Page Nos. 143 - 145
1. How does the force of gravitation between two objects change when the distance between them is reduced to half?
We know gravitational force between two objects –
F = GMm/R2, where G = gravitational constant, M = mass of the 1st object, m = mass of the 2nd object, r = distance between the objects. When the distance r is reduced to half, the new force –
F1 = GMm/(r/2)2 = 4GMm/r2 = 4F. So, the gravitational force becomes four times.
2. Gravitational force acts on all objects in proportion to their masses, Why then, a heavy object does not fall faster than a light object?
When there is no air, all objects fall on ground with constant acceleration. That acceleration due to gravity is g. The value of g is 9.8 ms-2. The gravitational force acting on an object is –
F = GMm/R2, where G = gravitational constant, M = mass of the earth, m = mass of the object, R = radius of the earth. Also –
F = GMm/R2 = mg ,
g = GM/ R2 = constant for a particular location on the surface of the earth. It does not depend on the mass of the object. So, all objects, be it heavy or light, fall on ground with same acceleration.
3. What is the magnitude of the gravitational force between the earth and a 1 kg object on it’s surface? (Mass of the earth = 6X1024 kg, radius of the earth = 6.4×104 m).
Gravitational force = F = G [(M x m)/R2 ]
Here, M = mass of the earth = 6×1024 kg, R = radius of the earth = 6.4×104 m, m = mass of the object = 1 kg, G = universal gravitation constant = 6.70 x 10-11 N m2 Kg-2.
So, F = G [(Mxm)/R2 ] = [(6.70 x 10-11 x 6 x 1024 x 1) / (6.4×106)2 ] N. = 9.8 N .
4. The earth & the moon are attracted to each other by gravitational force. Does the earth attract the moon with a force that is greater or smaller or the same as the force with which the moon attracts the earth. Why?
The universal law of gravitation states that two objects attract each other with equal force, but in opposite direction. So the attraction force by the Moon towards the Earth & the attraction force by the Earth towards the Moon are equal & opposite. This attraction force between the Earth & the Moon is ⇒ F = GMm/d2, where G = gravitational constant, M = mass of the Earth, m = mass of the Moon, d = distance between the Earth & the Moon. Therefore, the magnitude of F is same for Earth & Moon.
5. If the moon attracts the earth, why does the earth not move towards the moon?
As per Newton’s 3rd law of motion, the attraction force by the Moon towards the earth is equal to the attraction force by the Earth towards the Moon. But the Moon accelerates faster than the earth. Because the mass of the Earth is much greater than the mass of the Moon. So the earth does not move towards the Moon.
6. What happens to the force between two objects if ⇒
a) the mass of one object is doubled?
b) the distance between the objects is doubled & tripled?
c) the masses of both objects are doubled
We know, the gravitational force between two objects –
F = GMm/r2, where G = gravitational constant, M = mass of the 1st object, m = mass of the 2nd object, r = distance between the objects.
a. When the mass of any object is doubled (here mass of the 2nd object is doubled) , then new force, F1 = GMx2m/ r2 = 2GMm/ r2 = 2xF . So, the gravitational force is doubled.
b. When the distance r is doubled, the new force, F2 = GMm/ (2r)2 = GMm/ 4r2 = F/4. So, the gravitational becomes one-fourth.
c. When the mass of both objects is doubled, then new force, F3 = Gx2Mx2m/ r2 = 4GMm/ r2 = 4xF. So, the gravitational force becomes four times.
7. What is the importance of universal law of gravitation?
The universal law of gravitation is very important. It describes many important phenomena in science. For example,
1. It explains the motion of planets around the Sun.
2. It also explains the motion of Moon around the Earth.
3. In addition, it describes the attraction force that binds us to the Earth.
4. It explains the occurring of tides due to the motion of the Sun & Moon.
8. What is the acceleration of free fall?
When any object falls from the upper place towards ground, we may term it as free falling under the influence of the gravitational force. Any free falling involves some changes in velocity. Because, from the time of starting of falling to the time of touching of the earth, there are some changes in velocity. Also we know that any change in velocity is associated with acceleration. So free falling is actually an accelerated velocity. This acceleration is created by the gravity i.e. gravitational force. It is denoted by ɡ. The unit of ɡ is m/s2 or ms-2 i.e. same as that of acceleration. The value of ɡ is 9.8 ms-2.
9. What do we call the gravitational force between the earth & an object?
The gravitational force between the earth & an object is the of that object. This gravitational force (F) is equal to (m x ɡ) where ɡ is acceleration due to gravity. So, the weight of an object is ⇒
W = F = m x ɡ (m= mass of the object, ɡ = acceleration due to gravity). The S.I. Unit of weight is same as that of force and it is Newton.
10. 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].
We know, weight of an object – W = mg (m=mass of the object, g= acceleration due to gravity). Therefore,
weight at the poles – Wpole = m x gpole .
And weight at theequator – Wequator = m x gequator .
Given that, gpole > gequator . So, Wpole > Wequator . Hence, the friend will not agree. Because he will find less weight during measurement.
11. Why will a sheet of paper fall slower than one that is crumpled into a ball?
A sheet of paper is crumpled into a ball. As a result, its density increases. Therefore, resistance by the air against its motion decreases. So it fall faster than the sheet of paper.
12. Gravitational force on the surface of the moon is only 1/6 as strong as gravitational force on the earth. What is the weight of a 10 kg object on the moon and on the earth?
Weight of an object on earth = (m x g) N.
Here , m = mass of the object = 10 kg, g = acceleration due to gravity = 9.8 ms-2.
Therefore, weight on earth = (10 x 9.8) N = 98 N and weight on moon = (1/6) x 98 N = 16.3 N.
13. A ball is thrown vertically upwards with a velocity of 49 m/s. Calculate i) the maximum height to which it rises and ii) total time it takes to return to the surface of the earth.
i) The ball is thrown upwards i.e. against gravitational force. So ‘g’ i.e. acceleration due to gravity acts against velocity. Then –
⇒ 2 x (-g) x s = v2 – u2 [∵ (-) g indicates that g acts against velocity ]
⇒ (-)2 x 9.8 x s = 02 – 492 [∵s = displacement, v=final velocity=0, initial velocity=49m/s ]
So, s = 492 / 2×9.8 m = 122.5 m.
ii) Time to reach maximum height –
⇒ v = u – g x t [∵ (-) g indicates that g acts against velocity ]
⇒ t = (u –v)/g [v = final velocity = 0, initial velocity = 49 m/s]
Therefore, t = 49/9.8 s = 5 s.
Time to reach earth – when the ball starts falling after reaching maximum height, it’s initial velocity is zero and it falls due to gravitational force with acceleration g. So –
s = ut + (½)xg x t2 [ s= 122.5 m, u = 0, t=? ]
⇒ 122.5 = 0xt + (½)x9.8xt2
⇒ t2 = (122.5 x 2 ) / 9.8 = 25
Therefore, t = 5 s. So total time taken = (5 + 5) s = 10 s.
Time to reach earth – when the ball starts falling after reaching maximum height, it’s initial velocity is zero and it falls due to gravitational force with acceleration g. So –
s = ut + (½)xg x t2 [ s= 122.5 m, u = 0, t=? ]
⇒ 122.5 = 0xt + (½)x9.8xt2
⇒ t2 = (122.5 x 2 ) / 9.8 = 25
Therefore, t = 5 s. So total time taken = (5 + 5) s = 10 s.
14. A stone is released from the top of a tower of height 19.6 m. Calculate it’s final velocity just before touching the ground.
2g s = v2 – u2
⇒ 2 x 9.8 x 19.6 = v2 – 02 [ ∵ s = 19.6 m, u=0, v=? ]
⇒ v2 = 19.6 x 19.6 = 384.16
So, v = 19.6 m/s.
15. A stone is thrown vertically upward with an initial velocity of 40 m/s. Taking g =10 m/s2, find the maximum height reached by the stone. What is the net displacement & the total distance covered by the stone?
2g s = v2 – u2
⇒ 2x 10 x s = v2 – 02 [ ∵ s = ?, g = 10 m/s2, u = 0, v = 40 m/s ]
Therefore, s = 402 / (2 x 10) m = 80 m.
Maximum height reached = 80 m. After reaching maximum height 80 m, the stone falls back to ground again covers 80 m. So total distance covered = 80 + 80 m = 160 m. Basically displacement is the distance between starting point and ending point. As in this case, the stone returns back to starting point, so net displacement is zero.
16. Calculate the force of gravitation between the earth and the sun. Given data ⇒ mass of the earth = 6 x1024 kg, mass of the sun = 2 x 1030 kg, the average distance between the sun & the earth = 1.5 x 1011 m.
Gravitational force = F = G [(M x m)/d2]
Here, M = mass of the earth = 6X1024 kg, d= distance between the sun & the earth = 1.5 x 1011 m, m = mass of of the sun = 2 x 1030 kg, G = universal gravitation constant = 6.673 x 10-11 N m2 Kg-2.
So, F = G [ ( M x m ) / d2 ]
= [ (6.673 x 10-11 x 6 x 1024 x 2 x 1030) / (1.5 x 1011)2 ] N
= 35.58 x 1021 N = 3.56 x 1022 N.
17. A stone is allowed to fall from the top of a tower 100 m high and at the same time another stone is projected vertically upwards from the ground with a velocity of 25 m/s. Calculate when and where the two stones will meet.
Let assume both the stone meet after t seconds. Distance covered by 1st stone plus distance covered by 2nd stone will be 100 m. 1st stone fall in favour of g and 2nd stone moves against g. Distance covered by 1st stone-
s1 = ut + ½ xg x t2 [∵ u = 0]
⇒ s1 = 0xt + ½ x 9.8 x t2 = 4.9t2
Similarly, distance covered by 2nd stone s2 = ut – ½ xg x t2 [∵ u = 25 m/s]
⇒ s2 = 25xt – ½ x 9.8 x t2 = 25t – 4.9t2.
Now , s1 + s2 = 100
⇒ 9t2 + 25t – 4.9t2 = 100
⇒ 25 t = 100
So, t = 4 s.
Now, s1 = 4.9 x 42 m = 78.40 m and s2 = 25×4 – 4.9×42 = 21.6 m. They meet 78.40 m from top or 21.6 m from ground.
Here we have taken g = 9.8 m/s2. If you take g = 10 m/s2, the answer will be 80 m from top or 20 m from ground.
18. 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, c) it’s position after 4 s.
a) Total time taken = 6 s. So it takes 3 s to reach maximum height and 3 s to return back to the thrower. initial velocity = u = ?, final velocity = v = 0, g = 9.8 m/s2 . When stone goes up g acts against it. So,
⇒ v = u + (-g)xt ,
⇒ 0 = u – 9.8×3 ,
So, u = 29.4 m/s .
b) Maximum height it reaches = s = ut + ½ x(-g)xt2 = 29.4×3 + ½ x(-)9.8×32 = 44.1 m.
c) After 3 s it reaches maximum height, then it starts falling when initial speed = u = 0.
So, free falling in 1 s i.e. (4 s – 3 s) = s = ut + ½ x(g)xt2 = 0x1 + ½ x9.8×12 = 4.9 .
So after 4 s, it’s position is (44.1 – 4.9) m or 39.2 m up from the thrower.
19. In what direction does the buoyant force on an object immersed in a liquid act?
The buoyant force always acts in the upward direction. An object may sink or float in a liquid. But it always experiences buoyant force in the upward direction. That is also applicable in case of an object suspended in a liquid. The more you go deep into a liquid, the more buoyant force you experience in the upward direction.
20. Why does a block of plastic released under water come up to the surface of water?
In this case, two forces act on the block of plastic immersed in water. Firstly, the gravitational force pulls the block of plastic downwards. Secondly, the buoyant force pushes that object upwards. The buoyant force is greater than the gravitational force in this case. So the block of plastic comes up to the surface on the water as soon as it is released within water.
21. The volume of a 50g substance is 20 cm3. If the density of water is 1 g cm-3, will the substance float or sink?
Density of substance = (50/20) g cm-3 = 2.5 g cm-3 . As the density of the substance is greater than the density of the water, so the substance will sink.
22. The volume of a 500 g sealed packet is 350 cm3. Will the packet float or sink in water if the density of water is 1 g cm-3 ? What will be the mass of the water displaced by this packet?
Answer: Density of sealed packet = (500/350) = 1.42 g cm-3. Now we get that density of the sealed packet is greater than the density of water. So the substance will sink and displace water equal to it’s volume. So volume of displaced water is 350 cm3. Also given density of water is 1 g cm-3. We know-
Density = mass/volume , so mass = density x volume = 350 x 1 = 350 g.
Gravitation Science Chapter 10 – To Be Continued – Gravitation Science Chapter 10