Chapter 1 - Gravitation
1. Study the entries in the following table and rewrite them putting the connected items in a single row.
| I | II | III |
|---|---|---|
Mass | m/s² | Zero at the centre |
Weight | kg | Measure of inertia |
Acceleration due to gravity | Nm²/kg² | Same in the entire universe |
Gravitational constant | N | Depends on height |
Ans:
| I | II | III |
|---|---|---|
Mass | kg | Measure of inertia |
Weight | N | Zero at the centre |
Acceleration due to gravity | m/s² | Depends on height |
Gravitational constant | Nm²/kg² | Same in the entire universe |
2. 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 the same as their values on Mars? Why?
Ans: Difference between mass and weight of an object;
| Mass | Weight |
|---|---|
(i) Mass is the quantity of matter contained in an object. | (i) Weight is the force with which the earth attracts an object. |
(ii) Mass remains the same everywhere. | (ii) Weight of a body keeps changing from place to place. |
(iii) Mass is measured in kilogram (kg). | (iii) Weight is measured in newton (N). |
(iv) Mass is a scalar quantity. | (iv) Weight is a vector quantity. |
(v) Mass of an object can never be zero. | (v) Weight of an object becomes zero at the centre of the earth. |
(i) Mass is a fundamental quantity whose value remains the same everywhere. Hence, the mass of an object on earth will be the same as its value on Mars.
(ii) Weight of an object is a product of mass and gravitational acceleration, i.e., W = F = mq.
(iii) As the weight depends on the value of acceleration due to gravity (g) which changes from place-to-place, and is different for earth and Mars, the weight of the object on earth will be different than its value on Mars.
b. What are (i) free fall (ii) acceleration due to gravity (iii) escape velocity (iv) centripetal force?
Ans:
(i) Free fall:
The motion of any object under the influence of the force of gravity alone is called free fall. During free fall, the force of air also acts on an object. Thus real free fall is possible only in vacuum because there is no air.
For example:
a. A stone dropped down an empty well.
b. An apple falling from the tree.
(ii) Acceleration due to gravity:
The acceleration produced in a body under the influence of the force of gravity alone is called acceleration due to gravity. It’s S.I. unit is m/s². It is denoted by ‘g’ and its value at the surface of earth is 9.8 m/s².
(iii) Escape velocity:
The minimum velocity with which a body should be projected from the surface of a planet or moon, so that it escapes from the gravitational influence of the planet or moon is called escape velocity. The escape velocity from earth is about 11.186 km/s.
(iv) Centripetal force:
The force acting on any object moving in a circle and directed towards the centre of the circle is called centripetal force.
For example:
a. Planets orbiting around the sun.
b. The necessary centripetal force is provided by push due to rails on the wheels of the train during taking a turn.
c. Write the three laws given by Kepler. How did they help Newton to arrive at the inverse square law of gravity?
Ans:
(i) Kepler’s First Law – The Law of Orbits:
The orbit of a planet is an ellipse with the sun at one of the foci. This law is popularly known as the law of orbits.
(ii) Kepler’s second law – The law of equal areas:
The line joining the planet and the sun sweeps equal areas in equal intervals of time. This law is known as the law of areas.
(iii) Kepler’s Third Law – The Law of Periods:
The square of the orbital period of revolution of a planet around the Sun is directly proportional to the cube of the mean distance of the planet from the Sun. This is written as T² ∝ a³. This law is known as the law of Periods.
Derivation of inverse square law of gravity with the help of Kepler’s law:
(i) The centripetal force acting on a planet of mass m and velocity v, revolving at a distance of r from the Sun, is given as,
F = \(\large \frac {mv²}{r}\) …(i)
(ii) Let T be the period of revolution of the planet and 2πr be the distance travelled by the planet in one revolution. Then the speed of the planet is given as,
speed (v) = \(\large \frac {distance\, travelled}{time\, taken}\)
speed (v) = \(\large \frac {2πr}{T}\) …(ii)
(iii) Substituting equation (ii) in (i), we get,
F = \(\large \frac {m\, (\large \frac {2πr}{T})²}{r}\)
∴ F = \(\large \frac {4mπ²r²}{rT²}\)
∴ F = \(\large \frac {4mπ²r}{T²}\)
(iv) Multiplying and dividing by r², we get,
F = \(\large \frac {4mπ²r³}{r²T²}\)
∴ F = \(\large \frac {4mπ²}{r²}\) × \((\large \frac {r³}{T²})\) …(iii)
(v) But, According to Kepler’s third law,
T² ∝ r³
∴ \(\large \frac {T²}{r³}\) = constant (K) …(iv)
Substituting equation (iv) in (iii), we get,
F = \(\large \frac {4mπ²}{r²K}\)
But,
\(\large \frac {4mπ²}{K}\) = constant
∴ F ∝ \(\large \frac {1}{r²}\)
Thus, with the help of Kepler’s third law, Newton concluded that the centripetal force acting on the planet must be inversely proportional to the square of the distance between the planet and the Sun and hence, postulated the inverse square law of gravitation.
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 the same as the time taken to come down.
Ans:
(i) Consider a stone thrown vertically upwards with initial velocity ‘u’. It reaches a height ‘h’ before coming down.
(ii) The kinematical equations of motion are given as,
v = u + at …(i)
s = ut + \(\large \frac{1}{2}\) at² …(ii)
v² = u² + 2as …(iii)
iii. For upward motion of the stone,
a = – g …[negative sign indicates the direction of force is opposite to that of velocity]
v = 0 …[∵ at the highest point velocity becomes zero]
Substituting this in equation (i), the time t₁ taken by the stone to reach the maximum height is given as,
0 = u – gt₁
∴ gt₁ = u
∴ t₁ = \(\large \frac {u}{g}\) …(iv)
Similarly, substituting the values of a and v in equation (iii), the maximum height h which the stone reaches is given as,
0² – u² = – 2gh
∴ – u² = – 2gh
∴ u² = 2gh
∴ h = \(\large \frac {u²}{2g}\) …(v)
(iv) For downward motion of the stone,
a = g
u = 0 …[∵ At maximum height, initial velocity is zero]
Substituting this in equation (ii), the time t₂ taken by the stone to come down is given as,
h = 0 + \(\large \frac {1}{2}\) gt₂²
∴ h = \(\large \frac {1}{2}\) gt₂²
∴ t₂² = \(\large \frac {2h}{g}\)
∴ t₂ = \(\sqrt {\large \frac {2h}{g}}\) …(vi)
Substituting equation (v) in (vi), we get,
t₂ = \(\sqrt {\large \frac {2}{g} × \large \frac {u²}{2g}}\)
∴ t₂ = \(\sqrt {\large \frac {u²}{g²}}\)
∴ t₂ = \( \large \frac {u}{g}\) …(vii)
Thus, from equations (iv) and (vii), we can conclude that the time taken by the stone to go up is the 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?
Ans:
(i) The weight of an object is defined as the force with which the earth attracts the object. It is given as, W = F = mg
(ii) The weight of an object depends on the mass of the object and the value of acceleration due to gravity.
(iii) If the value of g doubles, the force with which the earth attracts the object also becomes twice.
(iv) Thus, the object becomes twice as heavier, making it harder to be pulled along the floor.
3. Explain why the value of g is zero at the centre of the earth.
Ans:
(i) The acceleration due to gravity (g) on earth’s surface is given as, g = \(\large \frac {GM}{R²}\). The value of g depends on the mass (M) of the earth and the radius (R) of the earth.
(ii) As we go inside the earth, our distance from the centre of the earth decreases and no longer remains equal to the radius of the earth (R).
(iii) Along-with the distance, the part of the earth which contributes towards the gravitational force also decreases, decreasing the value of M.
(iv) Due to the combined result of change in R and M, the value of g becomes zero at the centre of the earth.
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.
Ans:
(i) According to Kepler’s third law, the square of orbital period of revolution T of a planet around a star is directly proportional to the cube of the mean distance R of the planet from the star.
T² ∝ R³
∴ T² = k(R)³ …(i)
Where, k is constant of proportionality.
(ii) When the planet is at a distance of 2R from the star, then its period of revolution T’ will be,
T’² ∝ (2R)³
T’² = k(2R)³ …(ii)
Where, k is constant of proportionality.
(iii) Dividing equations (i) and (ii), we get,
\(\large \frac {T²}{T’²}\) = \(\large \frac {k(R)³}{k(2R)³}\)
∴ \(\large \frac {T²}{T’²}\) = \(\large \frac {R³}{8R³}\)
∴ \(\large \frac {T²}{T’²}\) = \(\large \frac {1}{8}\)
∴ T² = \(\large \frac {1}{8}\) T’²
∴ T’² = 8 T’²
∴ T’ = \(\sqrt {8}\) T
Thus, for a planet at a distance of 2R from the star, it’s period of revolution will be \(\sqrt {8}\) 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:
Time (t) = 5 s
Height (s) = 5 m
To find:
Gravitational acceleration (g)
Solution:
We know that,
s = ut + \(\large \frac {1}{2}\) at²
∴ 5 = 0t + \(\large \frac {1}{2}\) × g × 5²
∴ 5 = \(\large \frac {1}{2}\) × g × 25
∴ \(\large \frac {5\, ×\, 2}{25}\) = g
∴ \(\large \frac {2}{5}\) = g
∴ g = 0.4 m/s²
Ans: The gravitational acceleration of the planet is 0.4 m/s².
b. The radius of planet A is half the radius of planet B. If the mass of A is m\(_{A}\), what must be the mass of B so that the value of g on B is half that of its value on A?
Given:
For planet A:
mass = m\(_{A}\)
radius R\(_{A}\) = \(\large \frac {R_{B}}{2}\)
acceleration due to gravity = g\(_{A}\)
For planet B:
radius = R\(_{B}\)
acceleration due to gravity g\(_{B}\) = \(\large \frac {g_{A}}{2}\)
To find:
Mass of planet B, i.e. M\(_{B}\)
Solution:
We know that,
g = \(\large \frac {GM}{R²}\)
∴ M = \(\large \frac {gR²}{G}\)
∴ M\(_{A}\) = \(\large \frac {g_{A} R_{A}²}{G}\)
and M\(_{B}\) = \(\large \frac { g_{B} R_{B}²}{G}\)
Dividing M\(_{A}\) by M\(_{B}\),
\(\large \frac {M_{A}}{M_{A}}\) = \(\large \frac {\large \frac {g_{A} R_{A}²}{G}}{\large \frac {g_{B} R_{B}²}{G}}\)
∴ \(\large \frac {M_{A}}{M_{B}}\) = \(\large \frac {\large \frac {g_{A}(\large \frac {R_{B}}{2})²}{G}}{\large \frac {(\large \frac {g_{A}}{2})R_{B}²}{G}}\) …[Substituting the values of \(R_{A}\) and \(g_{B}\)]
∴ \(\large \frac {M_{A}}{M_{B}}\) = \(\large \frac {\large \frac {g_{A}\, ×\, R_{B}²}{4}}{\large \frac {g_{A}\, ×\, R_{B}²}{2}}\)
∴ \(\large \frac {M_{A}}{M_{B}}\) = \(\large \frac {2}{4}\) …\([\)∵ \(\large \frac {\large \frac {a}{b}}{\large \frac {c}{d}}\) = \(\large \frac {a\,×\,d}{b\,×\,c}]\)
∴ \(\large \frac {M_{A}}{M_{B}}\) = \(\large \frac {1}{2}\)
∴ \(M_{B}\) = \(2 M_{A}\)
Ans: The mass of planet B is \(2 M_{A}\).
c. The mass and weight of an object on Earth is 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 \(\large \frac {1}{6}\) of that on the earth.
Given:
Mass on earth (\(M_{e}\)) = 5 kg
Weight on earth (\(W_{e}\)) = 49 N
Acceleration due to gravity on moon (\(g_{M}\)) = \(\large \frac {9.8}{6}\) = 1.63 m/s²
To find:
Mass (\(M_{m}\)) on moon
Weight (\(W_{m}\)) on moon
Solution:
We know that,
\(W_{m}\) = \(M_{m}\) × \(g_{m}\)
The mass of the object is independent of gravity and remains unchanged i.e., \(M_{e}\) = \(M_{m}\) = 5 kg.
∴ \(W_{m}\) = 5 × 1.63
∴ \(W_{m}\) = 8.17 N
Ans: On the moon, the mass of the object is 5 kg and weight is 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 (s) = 500 m
Acceleration due to gravity (g) = 10 m/s²
To find:
(i) Initial velocity (u)
(ii) time taken (t)
Solution:
For upward motion of the ball, (v) = 0
a = – g = – 10m/s²
We know that,
v² = u² + 2as
∴ 0 = u² + 2(–10) × 500
∴ 0 = u² – 20 × 500
∴ 0 = u² – 10000
∴ u² = 10000
∴ u = 100 m/s
For downward motion of the ball, (u) = 0.
a = g = 10 m/s²
We know that,
s = ut + \(\large \frac {1}{2}\) at²
∴ 500 = 0 + \(\large \frac {1}{2}\) × 10t²
∴ 500 = 5t²
∴ \(\large \frac {500}{5}\) = t²
∴ t² = 100
∴ t = 10 s
Time for upward journey of the ball will be the same as time for downward journey i.e., 10 s.
∴ Total time taken = 2 × 10
∴ Total time taken = 20 s
Ans: The initial velocity of the object is 100 m/s and the total time taken by the object to reach the height and come down is 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:
Initial velocity (u) = 0
Acceleration due to gravity (g) = 10 m/s²
time (t) = 1s
To find:
Height of the table (s)
Velocity of the ball (v)
Solution:
We know that,
s = ut + \(\large \frac {1}{2}\) at²
∴ s = 0 × t + \(\large \frac {1}{2}\) × 10 × 1²
∴ s = \(\large \frac {1}{2}\) × 10
∴ s = 5m
We know that,
v² = u² + 2as
∴ v² = 0 + 2 × 10 × 5
∴ v² = 100
∴ v = 10 m/s
Ans: The velocity of the ball when it reaches the ground is 10 m/s, and the height of the table is 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 x 10⁻¹¹ Nm² kg⁻²
Given:
Mass of the earth (\(M_{e}\)) = 6 × 10²⁴ kg
Mass of the moon (\(M_{m}\)) = 7.4 × 10²² kg
Distance (r) = 3.84 × 10⁵ km
= 3.84 × 10⁵ × 10³ m,
= 3.84 × 10⁸ m,
Gravitational constant (G) = 6.7 x 10⁻¹¹ Nm² kg⁻²
To find:
Gravitational force (F)
Solution:
We know that,
F = \(\large \frac {Gm₁m₂}{R²}\)
∴ F = \(\large \frac {6.7\, ×\, 10⁻¹¹\, ×\, 6\, ×\, 10²⁴\, ×\, 7.4\, ×\, 10²²}{(3.84\, ×\, 10⁴)²}\)
∴ F = \(\large \frac {6.7\, ×\, 6\, ×\, 7.4\, ×\, 10⁻¹¹⁺²⁴⁺²²}{3.84\, ×\, 3.84\, ×\, 10¹⁶}\)
∴ F = \(\large \frac {297.48\, ×\, 10³⁵}{14.7456\, ×\, 10¹⁶}\)
∴ F = \(\large \frac {297.48\, ×\, 10³⁵}{14.7456\, ×\, 10¹⁶}\)
∴ F = 20.17 × 10¹⁹
∴ F = 2.017 × 10²⁰ N
Ans: The gravitational force between the earth and the moon is 2.017 × 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 x 10⁻¹¹ Nm² kg⁻²
Given:
Mass of the earth (\(M_{e}\)) = 6 × 10²⁴ kg
Gravitational force (F) = 3.5 × 10²² N
Distance (r) = 1.5 × 10¹¹ m
Gravitational constant (G) = 6.7 x 10⁻¹¹ Nm² kg⁻²
To find:
Mass of Sun (\(M_{s}\))
Solution:
We know that,
F = \(\large \frac {Gm₁m₂}{R²}\)
∴ \(M_{s}\) = \(\large \frac {Fr²}{GM_{e}}\)
∴ \(M_{s}\) = \(\large \frac {3.5\, ×\, 10²²\, ×\, (1.5\, ×\, 10¹¹)²}{6.7\, ×\, 10⁻¹¹\, ×\, 6\, ×\, 10²⁴}\)
∴ \(M_{s}\) = \(\large \frac {3.5\, ×\, 10²²\, ×\, 2.25\, ×\, 10²²}{40.2\, ×\, 10¹³}\)
∴ \(M_{s}\) = \(\large \frac {7.88\, ×\, 10⁴⁴}{40.2\, ×\, 10¹³}\)
∴ \(M_{s}\) = 1.96 × 10³⁰ kg
Ans: The mass of the Sun is 1.96 × 10³⁰ kg.