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Practice worksheet · Class 11 Physics

Gravitation

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Class 11 Physics · Gravitation

Gravitation worksheet for Class 11 Physics

35 practice questions on Gravitation for Class 11 Physics (Gravitation), from Priya Jain (M.Sc. Physics, B.Ed.). Open the worksheet to view, print or download it, or read the questions below.

31 MCQs 4 numericals Formula sheet Free PDF · 8 pages

Section A · Multiple Choice

Choose the correct option. Questions 1–31

  1. 1

    The value of the gravitational constant for two objects in vacuum is 6.67×10−11 N m2 kg−26.67 \times 10^{-11}\ \text{N m}^2\,\text{kg}^{-2}. Its value in a dense medium of density 101010^{10} g cm−3^{-3} will be:

    1. a 6.67×10−11 N m2 kg−26.67 \times 10^{-11}\ \text{N m}^2\,\text{kg}^{-2}
    2. b 6.67×10−10 N m2 kg−26.67 \times 10^{-10}\ \text{N m}^2\,\text{kg}^{-2}
    3. c 6.67×10−11 N m2 kg26.67 \times 10^{-11}\ \text{N m}^2\,\text{kg}^{2}
    4. d 6.67×10−21 N m2 kg−26.67 \times 10^{-21}\ \text{N m}^2\,\text{kg}^{-2}
  2. 2

    If the radius of the Earth shrinks by 1% but its mass remains the same, then the acceleration due to gravity on the Earth’s surface will:

    1. a Decrease
    2. b Remain the same
    3. c Increase
    4. d Cannot be determined
  3. 3

    At a certain place, the value of gg is 9.8 m/s². If a part of the Earth suddenly shrinks to half its size but the mass remains unchanged, then at that same place (assuming the distance rr of this place from the center of the Earth remains the same), the value of gg will be:

    1. a 4.9 m/s²
    2. b 9.8 m/s²
    3. c 3.1 m/s²
    4. d 19.6 m/s²
  4. 4

    The change in the value of gg at a height hh above the Earth’s surface is the same as the change at a depth xx inside the Earth (where both xx and hh are very small compared to the radius of the Earth). Then:

    1. a x=hx = h
    2. b x=2hx = 2h
    3. c x=h/2x = h/2
    4. d x=h2x = h^2
  5. 5

    Suppose a light planet revolves around a very massive star in a circular orbit of radius RR with a time period TT. If the gravitational force between the star and the planet is proportional to R−5/2R^{-5/2}, then:

    1. a T2∝R3T^2 \propto R^3
    2. b T2∝R3/2T^2 \propto R^{3/2}
    3. c T2∝R5/2T^2 \propto R^{5/2}
    4. d T2∝R7/2T^2 \propto R^{7/2}
  6. 6

    Two successive satellites of masses m1m_1 and m2m_2 are revolving in orbits of radii r1r_1 and r2r_2 respectively. If r1>r2r_1 > r_2, then which of the following relations regarding the speeds of the satellites v1v_1 and v2v_2 is true?

    1. a v1=v2v_1 = v_2
    2. b v1>v2v_1 > v_2
    3. c v1<v2v_1 < v_2
    4. d v1/r1=v2/r2v_1/r_1 = v_2/r_2
  7. 7

    The escape velocity for a rocket on Earth is 11.2 km/s. On a planet where the acceleration due to gravity is twice that of Earth and the diameter of the planet is twice the diameter of Earth, its value in km/s will be:

    1. a 11.2
    2. b 5.6
    3. c 22.4
    4. d 33.6
  8. 8

    The escape velocity from the surface of Earth is 11 km/s. If the radius of another planet is twice the radius of Earth but its mean density is the same as Earth’s, then the value of escape velocity from that planet will be:

    1. a 12 km/s
    2. b 22 km/s
    3. c 5.5 km/s
    4. d 15.5 km/s
  9. 9

    Two satellites A and B are revolving around a planet P in circular orbits of radii 4R4R and RR respectively. If the speed of satellite A is 3V3V, then the speed of satellite B will be:

    1. a 12V12V
    2. b 6V6V
    3. c 4V/34V/3
    4. d 3V/23V/2
  10. 10

    The radius of the Earth is 4000 miles. At a height of 10 miles above its surface, the change in acceleration due to gravity will be:

    1. a g/4000g/4000
    2. b g/400g/400
    3. c g/200g/200
    4. d g/100g/100
  11. 11

    If two objects of masses MM and mm revolve in circular paths of radii RR and rr respectively around their center of mass due to mutual gravitational attraction, then which of the following formulas will apply?

    1. a GMmR+r=mω2r\dfrac{GMm}{R + r} = m\omega^2 r
    2. b GMmR2=mω2r\dfrac{GMm}{R^2} = m\omega^2 r
    3. c GMm(R+r)2=mω2r\dfrac{GMm}{(R + r)^2} = m\omega^2 r
    4. d GMmR+r2=mω2r\dfrac{GMm}{R + r^2} = m\omega^2 r
  12. 12

    The escape velocity of an object of mass 1 kg on a planet is 100 m/s. The potential energy of the object on that planet will be:

    1. a −5000 J
    2. b −1000 J
    3. c −2400 J
    4. d −10000 J
  13. 13

    The height of a geostationary TV satellite will be (where TT is the duration of a day and RR is the radius of the Earth):

    1. a (4π2GMT2)1/3\left(\dfrac{4\pi^2 GM}{T^2}\right)^{1/3}
    2. b (4π2GMR2)1/3−R\left(\dfrac{4\pi^2 GM}{R^2}\right)^{1/3} - R
    3. c (GMT24π2)1/3−R\left(\dfrac{GMT^2}{4\pi^2}\right)^{1/3} - R
    4. d (GMT24π2)1/3+R\left(\dfrac{GMT^2}{4\pi^2}\right)^{1/3} + R
  14. 14

    Two objects of masses MM and mm (M>mM > m) revolve in circular orbits of radii RR and rr (r>Rr > R) about their center of mass. The angular velocity ω1\omega_1 of MM and the angular velocity ω2\omega_2 of mm are related as:

    1. a ω1>ω2\omega_1 > \omega_2
    2. b ω1<ω2\omega_1 < \omega_2
    3. c ω1=ω2\omega_1 = \omega_2
    4. d Mω1/R=mω2/rM\omega_1/R = m\omega_2/r
  15. 15

    The binding energy of the Earth–Moon system can be represented as:

    1. a GMeMm2r\dfrac{GM_eM_m}{2r}
    2. b −GMeMmr-\dfrac{GM_eM_m}{r}
    3. c GMeMmr\dfrac{GM_eM_m}{r}
    4. d −GMeMm2r-\dfrac{GM_eM_m}{2r}
  16. 16

    In planetary motion:

    1. a Rotational kinetic energy remains conserved
    2. b Angular momentum remains conserved
    3. c Linear kinetic energy remains conserved
    4. d Potential energy remains conserved
  17. 17

    If a planet of mass mm revolving around the Sun has an angular momentum LL, then the areal velocity (area swept per second by the line joining the planet to the Sun) will be:

    1. a mL2\dfrac{mL}{2}
    2. b 2mL2mL
    3. c L2m\dfrac{L}{2m}
    4. d Lm\dfrac{L}{m}
  18. 18

    A particle falls from infinity towards the Earth. Its velocity upon reaching the surface of the Earth will be:

    1. a Infinity
    2. b gR\sqrt{gR}
    3. c 2gR\sqrt{2gR}
    4. d gRgR
  19. 19

    If a person can jump 2 meters high on the surface of the Earth, then on a satellite where the acceleration due to gravity is 1.96 m/s², his jump height will be:

    1. a 5 m
    2. b 10 m
    3. c 20 m
    4. d 2 m
  20. 20

    The ratio of kinetic energy to potential energy of a satellite is:

    1. a 1 : 4
    2. b 4 : 1
    3. c 1 : 2
    4. d 2 : 1
  21. 21

    At a point located at a distance of 8000 km from the center of the Earth, the intensity of the gravitational field is 6 N/kg. The gravitational potential at that point is:

    1. a 8×1088 \times 10^{8}
    2. b 2.4×1022.4 \times 10^{2} J
    3. c 4.8×1074.8 \times 10^{7} J/kg
    4. d 6.4×10146.4 \times 10^{14} J
  22. 22

    The escape velocity of a planet is V0V_0. The value of escape velocity for a planet of the same density but twice the radius will be:

    1. a 0.5V00.5V_0
    2. b V0V_0
    3. c 2V02V_0
    4. d 4V04V_0
  23. 23

    The speed of a 500 kg satellite at a height of 10310^3 km from the Earth is 7.36×1037.36 \times 10^3 m/s. If the orbit of the satellite is circular, the gravitational force of the Earth on the satellite is:

    1. a 4×1024 \times 10^{2}
    2. b 3.66×1033.66 \times 10^{3}
    3. c 3.75×1043.75 \times 10^{4}
    4. d 5×1025 \times 10^{2}
  24. 24

    The minimum energy required to launch a 250 kg satellite from the surface of the Earth into an orbit at a height of 2Re2R_e will be:

    1. a 0.5×10100.5 \times 10^{10} J
    2. b 1.3×10101.3 \times 10^{10} J
    3. c 2.5×10102.5 \times 10^{10} J
    4. d 5×10105 \times 10^{10} J
  25. 25

    The ratio of the radii of two planets is pp and the ratio of their acceleration due to gravity is qq, then the ratio of the escape velocities of these planets will be:

    1. a (p/q)1/2(p/q)^{1/2}
    2. b (q/p)1/2(q/p)^{1/2}
    3. c pqpq
    4. d pq\sqrt{pq}
  26. 26

    Two bodies of masses M1M_1 and M2M_2 are dropped from the same height, then the ratio of their momenta at the surface of the Earth will be:

    1. a M1M2\sqrt{\dfrac{M_1}{M_2}}
    2. b M12M22\dfrac{M_1^2}{M_2^2}
    3. c M2M1\dfrac{M_2}{M_1}
    4. d M1M2\dfrac{M_1}{M_2}
  27. 27

    The percentage change in the weight of a body when moved from the equator to the poles on the surface of the Earth will be:

    1. a 4.5
    2. b 1.2
    3. c 0.43
    4. d 0.04
  28. 28

    What should be the rotational velocity of the Earth about its own axis so that the weight of a body at the equator becomes 3/53/5 of its present weight? (Radius of the Earth at the equator is RR.)

    1. a 5g2R\sqrt{\dfrac{5g}{2R}}
    2. b 2g5R\sqrt{\dfrac{2g}{5R}}
    3. c 3g5R\sqrt{\dfrac{3g}{5R}}
    4. d 5g3R\sqrt{\dfrac{5g}{3R}}
  29. 29

    On Earth, the pressure of a barometer and the height of the column are 10510^5 pascal and 760 mm respectively. If it is taken to the Moon, the height of the column will be:

    1. a Zero
    2. b 76 mm
    3. c 126 mm
    4. d 760 mm
  30. 30

    The escape velocity of a body from the Earth is 11.2 km/s. If it is projected at an angle of 45∘45^\circ with the horizontal, the value of the escape velocity will be:

    1. a 11.2 km/s
    2. b 22.4 km/s
    3. c 11.2/211.2/\sqrt{2} km/s
    4. d 11.2211.2\sqrt{2} km/s
  31. 31

    The figure shows the elliptical orbit of a planet mm relative to the Sun SS. The shaded area SCDSCD is twice the shaded area SABSAB. If the time taken by the planet to move from CC to DD is t1t_1 and the time taken to move from AA to BB is t2t_2, then:

    Elliptical orbit of planet m with the Sun S at one focus: a small shaded sector SAB on the near side and a long, narrow shaded sector SCD reaching the far end of the orbit
    1. a t1=t2t_1 = t_2
    2. b t1<t2t_1 < t_2
    3. c t1=4t2t_1 = 4t_2
    4. d t1=2t2t_1 = 2t_2

Section B · Short Answer & Numericals

Show all steps. Questions 32–35

  1. 32

    A mass MM is divided into two parts, xMxM and (1−x)M(1 - x)M. For a fixed distance between them, what should be the value of xx so that the gravitational force between the two parts is maximum?

  2. 33

    The escape velocity of Earth is 11.2 km/s. Compared to Earth, the radius and mass of Mars are approximately half and 1/10th respectively. What will the escape velocity from Mars be?

  3. 34

    Infinite masses, each of 1 kg, are placed along the x-axis at x=±1x = \pm 1 m, ±2\pm 2 m, ±4\pm 4 m, ±8\pm 8 m, ±16\pm 16 m, … What is the magnitude of the resultant gravitational potential at the origin (x=0x = 0), in terms of the gravitational constant GG?

  4. 35

    A rocket is fired vertically with a speed of 5 km/s from the Earth’s surface. How far from the Earth does the rocket go before returning to the Earth? Mass of the Earth =6×1024 kg= {6 \times 10^{24}\ \text{kg}}; mean radius of the Earth =6.4×106 m= {6.4 \times 10^{6}\ \text{m}}; G=6.67×10−11 N m2 kg−2G = {6.67 \times 10^{-11}\ \text{N m}^2\,\text{kg}^{-2}}.

Formula Sheet · Gravitation

Quick revision

Newton’s law of gravitation

Force between two masses F=Gm1m2r2F = \dfrac{G m_1 m_2}{r^2}
Gravitational constant G=6.67×10−11 m3kg⋅s2G = 6.67 \times 10^{-11}\ \dfrac{\text{m}^3}{\text{kg} \cdot \text{s}^2}

Acceleration due to gravity (gg)

At the surface g=GMR2=43πρGRg = \dfrac{GM}{R^2} = \dfrac{4}{3}\pi\rho G R
(i) Due to the shape of the Earth gegp=Rp2Re2,gp=ge+0.018 m/s2\dfrac{g_e}{g_p} = \dfrac{R_p^2}{R_e^2}, \quad g_p = g_e + 0.018\ \text{m/s}^2
(ii) At height hh gh=GM(R+h)2g_h = \dfrac{GM}{(R + h)^2}
At height hh, if h≪Rh \ll R gh=gs(1−2hR),Δgg=−2hRg_h = g_s\left(1 - \dfrac{2h}{R}\right), \quad \dfrac{\Delta g}{g} = -\dfrac{2h}{R}
(iii) At depth dd gd=GM(R−d)R3=gs(1−dR),Δgg=−dRg_d = \dfrac{GM(R - d)}{R^3} = g_s\left(1 - \dfrac{d}{R}\right), \quad \dfrac{\Delta g}{g} = -\dfrac{d}{R}

Effect of the Earth’s rotation on gg

(iv) At latitude λ\lambda g′=g−ω2Rcos⁡2λg' = g - \omega^2 R\cos^2\lambda
Poles and equator gp−ge=ω2R=0.034 m/s2g_p - g_e = \omega^2 R = 0.034\ \text{m/s}^2
Weightlessness at the equator (λ=0\lambda = 0) g′=0  ⇒  ω=gR=1.25×10−3 rad/sg' = 0 \;\Rightarrow\; \omega = \sqrt{\dfrac{g}{R}} = 1.25 \times 10^{-3}\ \text{rad/s}
Length of the day then T=1.40 h=5026.5 sT = 1.40\ \text{h} = 5026.5\ \text{s}
For weightlessness, the Earth must rotate 17 times faster17\ \text{times faster}

Variation of gg with distance rr from the Earth’s center

Inside (0<r≤R0 < r \le R) g=GMR3 rg = \dfrac{GM}{R^3}\,r
Outside (r>Rr > R) g=GMr2g = \dfrac{GM}{r^2}
Graph of g against r: g rises in a straight line from zero at the center to its surface value at r = R, then falls off as 1/r² outside the Earth

Gravitational potential: point mass & spherical shell

Due to a point mass V=−GMrV = -\dfrac{GM}{r}
Shell, outside (r>Rr > R) V=−GMrV = -\dfrac{GM}{r}
Shell, surface (r=Rr = R) V=−GMRV = -\dfrac{GM}{R}
Shell, inside (r<Rr < R) V=−GMRV = -\dfrac{GM}{R}
Graph of potential V against r for a spherical shell: constant at −GM/R from the center out to r = R, then rising towards zero as −GM/r

Potential due to a solid sphere

Outside (r>Rr > R) V=−GMrV = -\dfrac{GM}{r}
At the surface (r=Rr = R) V=−GMRV = -\dfrac{GM}{R}
Inside (r<Rr < R) V=−GM(3R2−r2)2R3V = -\dfrac{GM(3R^2 - r^2)}{2R^3}
At the center Vcenter=32 VsurfaceV_{\text{center}} = \dfrac{3}{2}\,V_{\text{surface}}
Graph of potential V against r for a solid sphere: −3GM/2R at the center, rising along a curve to −GM/R at r = R, then towards zero as −GM/r

Potential due to a thin ring

At a distance XX on the axis V=−GMR2+X2V = -\dfrac{GM}{\sqrt{R^2 + X^2}}
At the center V=−GMRV = -\dfrac{GM}{R}

Gravitational potential energy (UU)

Two point masses U=−GM1M2rU = -\dfrac{GM_1M_2}{r}
Uniform thin spherical shell U=−GM22RU = -\dfrac{GM^2}{2R}
Uniform solid sphere U=−35 GM2RU = -\dfrac{3}{5}\,\dfrac{GM^2}{R}
Change in P.E. ΔU=−GMm[1r2−1r1]\Delta U = -GMm\left[\dfrac{1}{r_2} - \dfrac{1}{r_1}\right]
At a height hh from the Earth’s surface Uh=−GMmR+h=−mgR1+hRU_h = -\dfrac{GMm}{R + h} = \dfrac{-mgR}{1 + \dfrac{h}{R}}
Work done against gravity W=mgh1+hRW = \dfrac{mgh}{1 + \dfrac{h}{R}}

Escape velocity (VeV_e)

Escape velocity Ve=2GMR=2Rg=R83πρGV_e = \sqrt{\dfrac{2GM}{R}} = \sqrt{2Rg} = R\sqrt{\dfrac{8}{3}\pi\rho G}
Maximum height attained by a body thrown with speed VV h=RVe2V2−1h = \dfrac{R}{\dfrac{V_e^2}{V^2} - 1}
If V>VeV > V_e (thrown faster than the escape velocity) V′=V2−Ve2V' = \sqrt{V^2 - V_e^2}

Orbital velocity & path of a satellite

For a nearby planet/satellite V0=GMR=Ve2V_0 = \sqrt{\dfrac{GM}{R}} = \dfrac{V_e}{\sqrt{2}}
Spiral path V<V0V < V_0
Circular path V=V0V = V_0
Elliptical path V0<V<VeV_0 < V < V_e
Parabolic path V=VeV = V_e
Hyperbolic path V>VeV > V_e
If F∝1/rnF \propto 1/r^n V∝1r(n−1)/2,T∝r(n+1)/2V \propto \dfrac{1}{r^{(n-1)/2}}, \quad T \propto r^{(n+1)/2}

Energies of a satellite

Potential energy U=−GMmr=−L2mr2U = -\dfrac{GMm}{r} = -\dfrac{L^2}{mr^2}
Kinetic energy K=GMm2r=L22mr2K = \dfrac{GMm}{2r} = \dfrac{L^2}{2mr^2}
Total (mechanical) energy E=−GMm2r=−L22mr2E = -\dfrac{GMm}{2r} = -\dfrac{L^2}{2mr^2}
Binding energy B.E.=−E=GMm2r\text{B.E.} = -E = \dfrac{GMm}{2r}
Energy to shift its orbit from R1R_1 to R2R_2 (change in total energy) ΔE=GMm2(1R1−1R2)\Delta E = \dfrac{GMm}{2}\left(\dfrac{1}{R_1} - \dfrac{1}{R_2}\right)
Graph of a satellite’s energies against orbit radius r: kinetic energy K positive, potential energy U negative and twice as large, total energy E (dashed) negative and between them, all approaching zero as r grows

To make a satellite escape from its orbit

Double its orbital kinetic energy K→2K(100% increase)K \to 2K \quad (100\%\ \text{increase})
Supply energy equal to its orbital energy (in magnitude) ∣E∣=GMm2r|E| = \dfrac{GMm}{2r}
Multiply its orbital velocity by 2\sqrt{2} V→2 V(41.4% increase)V \to \sqrt{2}\,V \quad (41.4\%\ \text{increase})

Kepler’s laws & time period of a satellite

1st law (law of orbits) Elliptical path, Sun at one focus\text{Elliptical path, Sun at one focus}
2nd law (law of areas): areal velocity dAdt=L2m=12 rv=constant\dfrac{dA}{dt} = \dfrac{L}{2m} = \tfrac{1}{2}\,rv = \text{constant}
3rd law (law of periods) T2∝a3,a=rmax⁡+rmin⁡2T^2 \propto a^3, \quad a = \dfrac{r_{\max} + r_{\min}}{2}
Velocity at aphelion Va=GMa(1−e1+e)V_a = \sqrt{\dfrac{GM}{a}\left(\dfrac{1 - e}{1 + e}\right)}
Velocity at perihelion Vp=GMa(1+e1−e)V_p = \sqrt{\dfrac{GM}{a}\left(\dfrac{1 + e}{1 - e}\right)}
Time period (basic) T=2πrVT = \dfrac{2\pi r}{V}
Time period (expanded) T=2πr3/2GM=2π(R+h)3gR2T = \dfrac{2\pi r^{3/2}}{\sqrt{GM}} = 2\pi\sqrt{\dfrac{(R + h)^3}{gR^2}}

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