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NCERT Class 11 Physics Chapter 7 Gravitation
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Gravitation
Chapter: 7
| Part – I |
EXERCISE
1. Answer the following:
(a) You can shield a charge from electrical forces by putting it inside a hollow conductor. Can you shield a body from the gravitational influence of nearby matter by putting it inside a hollow sphere or by some other means?
Ans: Gravitational forces are independent of the intervening medium. A body can not be sheltered from the gravitational influence of nearby matter.
(b) An astronaut inside a small spaceship orbiting around the earth cannot detect gravity. If the space station orbiting around the earth has a large size, can he hope to detect gravity?
Ans: The spaceship size is an important factor to feel the gravitational effect by the astronaut. If the size of the space station is very large, the magnitude of the gravity will also become appreciable and hence he can hope to detect it.
(c) If you compare the gravitational force on the earth due to the sun to that due to the moon, you would find that the Sun’s pull is greater than the moon’s pull. (you can check this yourself using the data available in the succeeding exercises). However, the tidal effect of the moon’s pull is greater than the tidal effect of the sun. Why?
Ans: The Moon, being closer, has a more significant difference in its gravitational pull on opposite sides of the Earth. This difference creates a stronger tidal force. The Sun, being farther away, has a smaller difference in its gravitational pull. Therefore, even though the Sun’s overall gravitational pull on the Earth is stronger, the Moon’s closer proximity gives it a more substantial tidal effect.
2. Choose the correct alternative:
(a) Acceleration due to gravity increases/decreases with increasing altitude.
Ans: As you move away from the Earth’s centre, the gravitational force decreases. This results in a decrease in acceleration due to gravity.
(b) Acceleration due to gravity increases/decreases with increasing depth (assume the earth to be a sphere of uniform density).
Ans: Acceleration decreases according to formula As gd = g (1- d/R)where d is depth and R is radius of earth.
(c) Acceleration due to gravity is independent of mass of the earth/mass of the body.
Ans: Acceleration due to gravity of body of mass m is given as, g = GM/ R2 where, universal gravitational constant is G, mass of earth is Mand radius of earth is R. This formula does not include the mass of the falling object, meaning that g is independent of the mass of the body.
(d) The formula –G Mm(1/r2 – 1/r1) is more/less accurate than the formula mg(r2 – r1) for the difference of potential energy between two points r2 and r1 distance away from the centre of the earth.
Ans: The difference in potential energy at radius r 1 and r 2 is given as, v =−GmM (1 r 2 − 1 r 1). Substituting the second equation in first, we get U ≈ mgh Hence, the first formula is more accurate.
3. Suppose there existed a planet that went around the Sun twice as fast as the earth. What would be its orbital size as compared to that of the earth?
Ans: TE = 1year, TNP = TE/2 = ½ year,
rE = 1 A.U., rNP = ?
Using Kepler’s third law, we have:
=
= 0.63 A.U.
4. Io, one of the satellites of Jupiter, has an orbital period of 1.769 days and the radius of the orbit is 4.22 × 108 m. Show that the mass of Jupiter is about one-thousandth that of the sun.
Ans: T2 = 4r2r3/GM
Where:
T is the orbital period.
R is the orbital radius.
G is the gravitational constant, 6.674 × 10-11 Nm2/ kg2
M is the mass of the central body.
Converting days to seconds:
T = 1.769 days × 24 hours/ day × 3600 seconds/hour
T = 1.769 × 86400 seconds
T = 152883 seconds
The radius of lo’s orbit is given as:
r = 4.22 × 108 m
Third law of Kepler’s
(152883s)2 = 4𝜋2 (4.22 × 108m)3
Solving for Mj (mass of jupiter):
Mj = 4𝜋2 (4.22 × 108 m)3 / G (152883)2
Substitute G = 6.674 × 10-11 Nm2/kg2
Mj = 4𝜋2 (4.22 × 108)3 / 6.674 × 10-11 (152883)2
Calculate the numerator:
4𝜋2 (4.22 × 108)3 ≈ 4 × 9.8696 × (7.52 × 1025) ≈ 2. 962 × 1027
Calculate the denominator:
6.674 × 10-11 × (152883)2 ≈ 6.674 × 10-11 × 2.336 × 1010 ≈
1.5559 × 100
Thus,
Mj = 2.962 × 1027 / 1.559
≈ 1.9 × 1027 kg
The mass of the sun Ms is approximately 1.989 × 1030 kg
Comparing Mj to Ms:
Mj / Ms
= 1.9 × 1027 / 1.989 × 1030
= 0.001.
5. Let us assume that our galaxy consists of 2.5 × 1011 stars each of one solar mass. How long will a star at a distance of 50,000 ly from the galactic centre take to complete one revolution? Take the diameter of the Milky Way to be 105 ly.
Ans: Here r = 50,000 ly
= 50,000 × 9.46 × 1015m
= 4.73 × 1020 m
M = 2.5 × 1011 Ansar mass
= 2.5 × 1011 × (2 × 1030) kg
We know that M = 4𝜋2r3 / GT2
T =
= 1.12 × 1016S.
6. Choose the correct alternative:
(a) If the zero of potential energy is at infinity, the total energy of an orbiting satellite is negative of its kinetic/potential energy.
Ans: Kinetic energy.
(b) The energy required to launch an orbiting satellite out of earth’s gravitational influence is more/less than the energy required to project a stationary object at the same height (as the satellite) out of earth’s influence.
Ans: Less.
7. Does the escape speed of a body from the earth depend on:
(a) The mass of the body.
Ans: The escape speed is independent of the mass of the body.
(b) The location from where it is projected.
Ans: The escape speed of a body depends upon the location.
(c) The direction of projection.
Ans: Yes, the escape speed depends on the height of the location.
(d) The height of the location from where the body is launched?
Ans: The escape velocity depends upon the gravitational potential at the point from which it is projected and this potential depends upon height also.
8. A comet orbits the sun in a highly elliptical orbit. Does the comet have a constant (a) linear speed, (b) angular speed, (c) angular momentum, (d) kinetic energy, (e) potential energy, (f) total energy throughout its orbit? Neglect any mass loss of the comet when it comes very close to the Sun.
Ans: Angular momentum and total energy do not vary throughout the orbit, whereas rest, all quantities vary in the orbit.
9. Which of the following symptoms is likely to afflict an astronaut in space:
(a) swollen feet.
(b) swollen face.
(c) headache.
(d) orientational problem.
Ans: The astronaut in space will suffer from: swollen face, headache and orientational problem.
10. In the following two exercises, choose the correct answer from among the given ones: The gravitational intensity at the centre of a hemispherical shell of uniform mass density has the direction indicated by the arrow (see Fig 7.11) (i) a, (ii) b, (iii) c, (iv) 0.
Ans: Intensity inside the shell is zero, so it will zero at P and Q also (Gravitational potential is constant).
11. For the above problem, the direction of the gravitational intensity at an arbitrary point P is indicated by the arrow (i) d, (ii) e, (iii) f, (iv) g.
Ans: As in the image the upper portion of the shell is missing, so the gravitational intensity at P and Q should act along (e) and (c) respectively.
12. A rocket is fired from the earth towards the sun. At what distance from the earth’s centre is the gravitational force on the rocket zero? Mass of the sun = 2×1030 kg, mass of the earth = 6×1024 kg. Neglect the effect of other planets etc. (orbital radius = 1.5 × 1011 m).
Ans: Given that:
Mass of the sun, Ms = 2 × 1030 kg
Mass of the earth, Me = 6 × 1024 kg
Distance from the earth to the sun, R = 1.5 × 1011m
Apply the formula:
x = 2.6 × 108m.
13. How will you ‘weigh the sun’, that is to estimate its mass? The mean orbital radius of the earth around the sun is 1.5 × 108 km.
Ans: Here r = 50,000 ly
= 50,000 × 9.46 × 1015m
= 4.73 × 1020 m
M = 2.5 × 1011 Ansar mass
= 2.5 × 1011 × (2 × 1030) kg
We know that M = 4𝜋2r3 / GT2
Radius R = 1.5 × 1011 m
Period T = 3.156 × 107 s
Gravitational constant G = 6.674 x 1011 m3 kg I s 2
Ms = 2 × 1030 kg.
14. A saturn year is 29.5 times the earth year. How far is Saturn from the sun if the earth is 1.50 × 108 km away from the sun?
Ans: We know that:
(T1 / T2) = (R1 / R2)3
Here T1 and T2 are the time periods of the two planets.
R1 and R2 are the distance of two planets from the sun.
Using Kepler’s Third Law:
(T₁/29.5T₁)² = (1.50 × 10⁸ km / R₂)³
1/29.5² = (1.50 × 10⁸ km / R₂)³
R₂³ = (29.5)² × (1.50 × 10⁸ km)³
R₂ = (29.5) (2/3) × 1.50 × 10⁸ km
R2 = 1.43 × 109 km.
15. A body weighs 63 N on the surface of the earth. What is the gravitational force on it due to the earth at a height equal to half the radius of the earth?
Ans: Here:
F Surface = GMm / R2
Given that:
F surface = 63 N
Fsurface = m × g
r = R + h
R = R/2 = 3R/ 2
New height:
Fheight =
F height = 4/9 × F surface
Fheight = 4/9 × 63 N
= 252 / 9
= 28 N.
16. Assuming the earth to be a sphere of uniform mass density, how much would a body weigh half way down to the centre of the earth if it weighed 250 N on the surface?
Ans: Weight of a body of mass m at the Earth’s surface, W =
mg = 250 N
Body of mass m is located at depth, d = ½ Re
Where
Re = Radius of Earth.
As gd = g
= mg d = mg
Here, d = R/2
Hence, mg = (250) ×
= 250 × ½ = 125 N.
17. A rocket is fired vertically with a speed of 5 km s-1 from the earth’s surface. How far from the earth does the rocket go before returning to the earth? Mass of the earth = 6.0 × 1024 kg; mean radius of the earth = 6.4 × 106 m; G = 6.67 × 10–11 N m2 kg–2 .
Ans: Initial speed of the rocket, v0 = 5km/s = 5000 m/s
Mass of the earth M = 6.0 × 1024 kg
Radius of the earth, R = 6.4 × 106 m
Gravitational constant, G = 6.67 × 10-11 N m2 kg-2
Initial kinetic energy of the rocket
= ½ mv2
= ½ x m × (5000)2
= 1.25 × 107 mj.
At distance “r” from the centre of the earth K.E. becomes zero:
The change in K.E.
= 1.25 × 107 – 0
= 1.25 × 107 mj.
This energy changes into potential energy.
Initial P.E. at the surface of earth: -GM e m / r
Final P.E. at distance
Change in P.E. = 6.25 × 107 – 4 × 104 m/r
Using law of conservation of energy,
= 6.25 × 107 m – 4 × 104m / r
= 1.25 × 107 m.
= i.e., r = 4 × 104 / 5 × 107
= 8 × 106 m.
18. The escape speed of a projectile on the earth’s surface is 11.2 km s–1. A body is projected out with thrice this speed. What is the speed of the body far away from the earth? Ignore the presence of the sun and other planets.
Ans: If v and v’ are the initial and final speeds of the body, its kinetic energies are ½ mv2 and ½ mv2 respectively. The initial potential energy of the body = -GM/R, and the final potential energy (at ∞) = 0.
By using the principle of conservation of energy, we get, Initial K.E. + Initial P.E. = Final K.E. + Final P.E.
= v’ = √8ve
= √ 8 × 11.2
= 31.68 kms-1.
19. A satellite orbits the earth at a height of 400 km above the surface. How much energy must be expended to rocket the satellite out of the earth’s gravitational influence? Mass of the satellite = 200 kg; mass of the earth = 6.0 × 1024 kg; radius of the earth = 6.4 × 106 m; G = 6.67 × 10–11 N m2 kg–2.
Ans: Given:
Height of the satellite above Earth’s surface, h = 400 km = 400,000 m
Mass of the satellite, m = 200 kg
Mass of the Earth, M = 6.0×1024 kg
Radius of the Earth, R = 6.4×106 m
Gravitational constant, G = 6.67×10−11 N m2 kg−2
= GMm / 2 (R + x)
= – 5.89 × 109 J
20. Two stars each of one solar mass (= 2×1030 kg) are approaching each other for a head on collision. When they are a distance 109 km, their speeds are negligible. What is the speed with which they collide? The radius of each star is 104 km. Assume the stars to remain undistorted until they collide. (Use the known value of G).
Ans: Given that:
Mass of each star, M = 2 × 1030 kg
Initial separation, r = 109 km = 1012 m
Gravitational constant, G = 6.67 × 10-11 Nm2/kg2
Initial potential energy:
U = – G × M × M / r
Final kinetic energy (K):
K = 0.5 × M × v2 + 0.5 × M × v2 = M × v2
Conservation of energy:
Initial potential energy = Final kinetic energy
– G × M × M / r = M × v2
Solving for v:
v = sqrt(- 2 × G × M / r)
Substituting the given values:
v = sqrt (- 2 × 6.67 × 10-11 Nm2/kg2 × 2 × 1030 kg / 1012 m)
v = sqrt (2.668 × 108 m2/s2)
v = 2.58 × 106 m/s
21. Two heavy spheres each of mass 100 kg and radius 0.10 m are placed 1.0 m apart on a horizontal table. What is the gravitational force and potential at the mid point of the line joining the centres of the spheres? Is an object placed at that point in equilibrium? If so, is the equilibrium stable or unstable?
Ans: Here G = 6.67 × 10-11 Nm2 Kg-2
M – 100 kg; R = 0.1m
Distance between two spheres, d = 1.0m.
Suppose that the distance of either sphere from the mid-point of the line joining their centre is r.
Then r = d/2 = 0.5 m.
The gravitational field at the mid- point due to two spheres will be equal and opposite.
Hence the resultant gravitational field at the mid – point = 0
The gravitational potential at the mid – point
= – 2.668 × 10-8 kg-1.
As the net force on the body placed at mid-point is zero, so the body is in equilibrium. If the body is displaced a little towards either mass body from its equilibrium position, it will not return back to its initial position of equilibrium. Hence, the equilibrium attained is unstable.
