Class 10 Science 1 Chapter 1 Gravitation – Textbook Exercise Questions and Answers | Maharashtra Board
Lesson 1: Gravitation
Class 10 Science
Q. 2. Answer the following question.
Q.a. What is the difference between mass and weight of an object? Will the mass and weight of an object on the Earth be the same as their values on Mars? Why?
Answer:
The following are the differences between the mass and weight of an object.
| MASS | WEIGHT |
|---|---|
| 1. Mass is the amount of matter present in an object. | 1. Weight of an object is the gravitational force with which a planet attracts the object. |
| 2. The SI unit of mass is kilogram (kg). | 2. The SI unit of weight is Newton (N). |
| 3. Its value remains the same everywhere. | 3. Weight of an object changes from place to place. |
| 4. A beam balance is used to measure the mass of an object. | 4. A spring balance is used to measure the weight of an object. |
| 5. Mass of an object cannot be zero. | 5. Weight of an object can be zero. |
i. Since mass is the amount of matter present in a body, and the amount of matter remains the same everywhere, the mass of an object on Earth will be the same as its mass on Mars.
ii. Whereas, the weight of an object on Earth and Mars will be different, as the acceleration due to gravity (g) is different for Earth and Mars.
Weight = Mass × g
Therefore, the mass remains the same, but the weight changes from Earth to Mars.
(i) Free Fall:
Whenever an object moves under the influence of gravitational force alone, it is said to be in free fall.
During free fall, the object may have any initial velocity. The important condition is that only gravitational force acts on the object.
Examples:
a) When an object is dropped from the top of a table, it falls down due to gravitational force and hence it is in free fall, if air resistance is neglected.
b) An apple falling from a tree is an example of free fall when air resistance is neglected.
(ii) Acceleration due to Gravity:
The acceleration gained by an object due to the gravitational force acting on it is called acceleration due to gravity. It is denoted by g.
Its SI unit is m/s² and its value near the surface of the Earth is approximately 9.8 m/s².
It is a vector quantity and is directed towards the centre of the Earth. The value of acceleration due to gravity changes from place to place. For example, on the Moon, the value of g is approximately one-sixth of its value on Earth.
(iii) Escape Velocity:
The minimum initial velocity with which an object must be projected from the surface of the Earth so that it escapes from the Earth's gravitational field and does not return to the Earth is called escape velocity.
The escape velocity from the Earth is approximately 11.2 km/s.
The escape velocity depends on the mass and radius of the planet. Spacecraft and rockets need sufficient velocity to overcome the gravitational attraction of the Earth.
(iv) Centripetal Force:
The force that acts on an object moving in a circular path and is directed towards the centre of the circular path is called centripetal force.
Fc = mv²/r
Here, m is the mass of the object, v is its velocity and r is the radius of the circular path.
The centripetal force is always directed towards the centre of the circular path.
Example: When a stone tied to a string is whirled in a circle, the tension in the string provides the centripetal force acting towards the centre.
Question
c. Write the three laws given by Kepler. How did they help Newton to arrive at the inverse square law of gravity?
Answer
Kepler gave the following three laws of planetary motion:
1. Law of Orbits
Every planet moves around the Sun in an elliptical orbit, with the Sun at one of the foci of the ellipse.
2. Law of Areas
The line joining a planet and the Sun sweeps out equal areas in equal intervals of time. Therefore, a planet moves faster when it is closer to the Sun and slower when it is farther away.
Diagram: The orbit of a planet moving around the Sun
3. Law of Periods
The square of the time period (T²) of a planet is proportional to the cube of the semi-major axis (r³) of its orbit.
How did Kepler's laws help Newton?
Kepler's second law showed that the force acting on a planet is directed towards the Sun. Thus, the force is a central force.
From Kepler's third law:
Newton used the formula for centripetal acceleration:
Therefore,
Since F = ma, Newton concluded that the gravitational force is inversely proportional to the square of the distance between the two bodies:
Thus, Kepler's laws provided the observational basis that helped Newton formulate the inverse square law of gravitation.
Given:
Initial velocity = u
Maximum height = h
Acceleration due to gravity = g
1. Time taken to go up
Using the first equation of motion:
v = u − gt
At the highest point, final velocity v = 0.
0 = u − gtup
tup = u/g
2. Maximum height reached
Using the third equation of motion:
v² = u² − 2gh
At the highest point, v = 0.
0 = u² − 2gh
h = u²/2g
3. Time taken to come down
During downward motion, initial velocity is 0.
Using the second equation of motion:
h = ½gtdown²
Substituting h = u²/2g:
u²/2g = ½gtdown²
u² = g²tdown²
tdown = u/g
Therefore,
tup = tdown
Hence, the time taken by the stone to go up is equal to the time taken to come down.
Answer:
The frictional force acting on a heavy object is given by:
F = μN
For an object lying on a horizontal floor, the normal reaction is:
N = mg
Therefore,
F = μmg
If g becomes twice
If the value of g becomes twice its original value, then:
g' = 2g
The new frictional force will be:
F' = μm(2g)
F' = 2μmg
F' = 2F
Therefore,
When g becomes twice, the normal reaction and hence the frictional force also become twice.
Therefore, it becomes twice as difficult to pull the heavy object along the floor.
3. Explain why the value of g is zero at the centre of the Earth.
Answer:
At the centre of the Earth, the gravitational force acting on an object is zero.
This is because the object is attracted equally in all directions by the mass of the Earth surrounding it. These gravitational forces cancel each other.
Therefore, the net gravitational force acting on the object is zero.
g = F/m
Since F = 0,
g = 0
Therefore,
The value of g is zero at the centre of the Earth because the gravitational forces acting in all directions cancel each other.
4. Let the period of revolution of a planet at a distance R from a star be T. Prove that if it was at a distance of 2R from the star, its period of revolution will be √8 T.
Given:
Distance of the planet from the star = R
Period of revolution = T
New distance from the star = 2R
Let the new period of revolution be T′.
Using Kepler's Third Law:
According to Kepler's third law,
T² ∝ R³
Therefore,
T′² / T² = (2R)³ / R³
T′² / T² = 8R³ / R³
T′² / T² = 8
Taking square root:
T′ / T = √8
T′ = √8 T
Therefore,
The period of revolution at a distance 2R is √8 T.
5. Solve the following examples.
(a) An object takes 5 s to reach the ground from a height of 5 m on a planet. What is the value of g on the planet?
Given:
Height, h = 5 m
Time, t = 5 s
Initial velocity, u = 0
Using the equation:
s = ut + 1⁄2gt2
5 = 0 + 1⁄2 × g × (5)2
5 = 25g⁄2
g = 10⁄25 = 0.4 m/s2
Answer: g = 0.4 m/s2
(b) The radius of planet A is half the radius of planet B. If the mass of A is MA, what must be the mass of B so that the value of g on B is half that of its value on A?
Given:
RA = RB⁄2
Therefore, RB = 2RA
gB = 1⁄2gA
We know:
g = GM⁄R2
gB⁄gA = MB⁄MA × RA2⁄RB2
1⁄2 = MB⁄MA × RA2⁄(2RA)2
1⁄2 = MB⁄MA × 1⁄4
MB = 2MA
Answer: MB = 2MA
(c) The mass and weight of an object on earth are 5 kg and 49 N respectively. What will be their values on the moon? Assume that the acceleration due to gravity on the moon is 1/6th of that on the earth.
Given:
Mass on Earth = 5 kg
Weight on Earth = 49 N
Mass remains the same on the Moon.
Therefore, Mass on Moon = 5 kg
Acceleration due to gravity on Moon:
gM = gE⁄6
Weight on Moon:
WM = mgM
WM = 5 × 9.8⁄6
WM = 49⁄6
WM = 8.17 N
Answer: Mass = 5 kg and Weight = 8.17 N
(d) An object thrown vertically upwards reaches a height of 500 m. What was its initial velocity? How long will the object take to come back to the earth? Assume g = 10 m/s2.
Given:
Height, h = 500 m
Acceleration due to gravity, g = 10 m/s2
Final velocity at maximum height, v = 0
Using:
v2 = u2 - 2gh
0 = u2 - 2 × 10 × 500
u2 = 10000
u = 100 m/s
Time taken to reach maximum height:
v = u - gt
0 = 100 - 10t
t = 10 s
Total time to come back to the earth:
T = 2t = 2 × 10 = 20 s
Answer: Initial velocity = 100 m/s and time taken = 20 s.
(e) A ball falls off a table and reaches the ground in 1 s. Assuming g = 10 m/s2, calculate its speed on reaching the ground and the height of the table.
Given:
Time, t = 1 s
Acceleration due to gravity, g = 10 m/s2
Initial velocity, u = 0
Speed on reaching the ground:
v = u + gt
v = 0 + 10 × 1
v = 10 m/s
Height of the table:
s = ut + 1⁄2gt2
s = 0 + 1⁄2 × 10 × (1)2
s = 5 m
Answer: Speed = 10 m/s and height of table = 5 m.
(f) The masses of the earth and moon are 6 × 1024 kg and 7.4 × 1022 kg, respectively. The distance between them is 3.84 × 105 km. Calculate the gravitational force of attraction between the two. Use G = 6.7 × 10-11 N m2 kg-2.
Given:
ME = 6 × 1024 kg
MM = 7.4 × 1022 kg
Distance = 3.84 × 105 km
G = 6.7 × 10-11 N m2 kg-2
Convert distance into metre:
r = 3.84 × 105 × 103
r = 3.84 × 108 m
Using Newton's law of gravitation:
F = GMEMM⁄r2
F = (6.7 × 10-11)(6 × 1024)(7.4 × 1022) ⁄ (3.84 × 108)2
F ≈ 2 × 1020 N
Answer: Gravitational force = 2 × 1020 N.
(g) The mass of the earth is 6 × 1024 kg. The distance between the earth and the Sun is 1.5 × 1011 m. If the gravitational force between the two is 3.5 × 1022 N, what is the mass of the Sun? Use G = 6.7 × 10-11 N m2 kg-2.
Given:
ME = 6 × 1024 kg
r = 1.5 × 1011 m
F = 3.5 × 1022 N
G = 6.7 × 10-11 N m2 kg-2
Using Newton's law of gravitation:
F = GMEMS⁄r2
Therefore,
MS = Fr2 ⁄ GME
MS = (3.5 × 1022)(1.5 × 1011)2 ⁄ (6.7 × 10-11)(6 × 1024)
MS ≈ 1.96 × 1030 kg
Answer: Mass of the Sun = 1.96 × 1030 kg.
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