Physics · Gravitation · chapter 7 of 14

Class 11 Physics Chapter 7: Gravitation worksheet with answer key

A 20-mark worksheet on Gravitation for about 40 minutes of class or homework: 5 multiple-choice, 3 very short answer and 3 short answer questions, written from the NCERT chapter and checked by a second reviewer. Print the PDF for the class; the answer key carries the marking scheme. Teachers may photocopy it freely.

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Question 1
Two small spheres attract each other with a gravitational force F. If the mass of each sphere is doubled and the distance between their centres is tripled, the new force is
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Answer: (a) 49 F

F ∝ m₁m₂/r². The product of masses becomes 2 × 2 = 4 times and r² becomes 9 times, so F′ = (4/9)F. 2/3 F treats the distance as entering to the first power; 9/4 F inverts the ratio.

Question 2
A satellite in a circular orbit around the Earth has kinetic energy 5.0 × 10⁹ J. Taking the potential energy to be zero at infinity, its total energy is
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Answer: (d) −5.0 × 10⁹ J

In a circular orbit P.E. : K.E. : E = −2 : 1 : −1, so E = −K.E. = −5.0 × 10⁹ J. −1.0 × 10¹⁰ J is the potential energy, not the total.

Question 3
An astronaut inside a satellite orbiting the Earth feels weightless because
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Answer: (a) the astronaut and the satellite are both in free fall towards the Earth

Gravity at the orbit is not small, and gravitational shielding is not possible. Both the astronaut and the satellite fall freely towards the Earth with the same acceleration, so the floor presses nothing on the astronaut.

Question 4
Treating the Earth as a uniform sphere, the acceleration due to gravity is greatest
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Answer: (b) on the surface of the Earth

Inside a uniform Earth g grows in proportion to r and is zero at the centre; outside it falls as 1/r². So g is greatest at r = RE, on the surface, and decreases whether one goes up or down.

Question 5
The dimensional formula of the universal gravitational constant G is
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Answer: (b) [M⁻¹ L³ T⁻²]

G = Fr²/(m₁m₂), so [G] = [M L T⁻²][L²]/[M²] = [M⁻¹ L³ T⁻²]. [M L² T⁻²] is the dimension of energy.

Question 6 · 2 marks
Starting from W(r) = −GME m/r, show that the gain in potential energy when a body of mass m is raised from the Earth's surface to a small height h is approximately mgh.
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ΔW = −GME m/(RE + h) + GME m/RE = GME m h/[RE(RE + h)]. For h << RE, RE + h ≈ RE, so ΔW ≈ (GME/RE²)mh = mgh, using g = GME/RE².

  • ΔW = GME m h/[RE(RE + h)] (1)
  • h << RE and g = GME/RE² give mgh (1)
Question 7 · 2 marks
Two point masses A and B are fixed 12 m apart. A small body placed on the line AB at 3 m from A experiences no net gravitational force. Find the ratio mB/mA.
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At the point, GmA m/3² = GmB m/9², since the body is 3 m from A and 12 − 3 = 9 m from B. So mB/mA = 81/9 = 9.

  • equates the two attractions with distances 3 m and 9 m (1)
  • mB/mA = 9 (1)
Question 8 · 2 marks
A 100 kg satellite moves around the Earth with angular momentum 4.0 × 10¹² kg m² s⁻¹ about the Earth's centre. Find the area swept out per second by the line joining the satellite to the Earth's centre. Does this rate change as the satellite moves along an elliptical orbit?
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ΔA/Δt = L/2m = 4.0 × 10¹²/(2 × 100) = 2.0 × 10¹⁰ m² s⁻¹. Gravity is a central force, so L is constant and the areal speed does not change along the orbit.

  • ΔA/Δt = L/2m = 2.0 × 10¹⁰ m² s⁻¹ (1)
  • constant, since L is conserved under a central force (1)
Question 9 · 3 marks
A 1 kg particle is at the origin. A 4 kg mass is fixed at (2 m, 0) and a 9 kg mass at (0, 3 m). Find the magnitude and direction of the resultant gravitational force on the 1 kg particle. Take G = 6.67 × 10⁻¹¹ N m² kg⁻².
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From the 4 kg mass: F₁ = G × 4 × 1/2² = G = 6.67 × 10⁻¹¹ N along +x. From the 9 kg mass: F₂ = G × 9 × 1/3² = G along +y. By vector addition, F = √2 G = 9.43 × 10⁻¹¹ N, directed at 45° to the +x axis, between the two masses.

  • F₁ = G along +x (1)
  • F₂ = G along +y (1)
  • resultant √2 G ≈ 9.4 × 10⁻¹¹ N at 45° (1)
Question 10 · 3 marks
Show that the period of a satellite in a circular orbit of radius r around the Earth varies as r3/2. A satellite close to the surface has period T₀. At what height above the surface (RE = 6400 km) is the period 2√2 T₀?
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Gravity provides the centripetal force: GME m/r² = mV²/r, so V = √(GME/r), and T = 2πr/V = 2πr3/2/√(GME), so T ∝ r3/2. For T/T₀ = 2√2 = 23/2, r/RE = 2, so r = 2RE and h = RE = 6400 km.

  • V² = GME/r from gravity as centripetal force (1)
  • T = 2πr3/2/√(GME) (1)
  • r = 2RE, h = 6400 km (1)
Question 11 · 3 marks
Show that the areal velocity of a planet about the Sun is L/2m, where L is its angular momentum and m its mass. Hence explain why the law of areas holds for any central force.
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In time Δt the radius vector sweeps a triangle of area ΔA⃗ = ½(r⃗ × v⃗Δt). So ΔA⃗/Δt = ½(r⃗ × p⃗)/m = L⃗/2m. A central force acts along r⃗, so its torque r⃗ × F⃗ = 0 and L⃗ is constant. Hence ΔA/Δt is constant for any central force, not only the inverse square force.

  • ΔA = ½ r × vΔt (1)
  • ΔA/Δt = L/2m (1)
  • central force: zero torque, L constant, so equal areas (1)
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