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Gravitation flash cards

Master Gravitation through 97 NEET-level recall cards, systematically structured one idea at a time. Revise concept-wise, identify the areas where you need improvement, and focus your preparation with greater precision.

Gravitation, question and answer

25 of this chapter's 97 cards, laid out open so you can read straight through. The remaining 72 are in the interactive deck, where the answer stays hidden until you commit to one.

  1. 1.State Newton's law of universal gravitation.

    Every particle attracts every other particle with a force directly proportional to the product of their masses and inversely proportional to the square of the distance between them: F=Gm1m2r2F=\dfrac{Gm_1m_2}{r^2}, directed along the line joining them.

    Hint: Product of masses over distance squared.

  2. 2.What is the value and SI unit of the universal gravitational constant GG?

    G=6.67×1011 N m2kg2G=6.67\times10^{-11}\ \text{N m}^2\,\text{kg}^{-2}. It is a scalar and a universal constant, independent of the medium.

    Hint: Order 101110^{-11}; units N m² kg⁻².

  3. 3.What are the dimensions of GG?

    [G]=[M1L3T2][G]=[M^{-1}L^{3}T^{-2}], obtained from G=Fr2m1m2G=\dfrac{Fr^2}{m_1m_2}.

    Hint: Rearrange F=Gm1m2/r2F=Gm_1m_2/r^2.

  4. 4.Is gravitational force a central force and a conservative force?

    Yes. It acts along the line joining the two masses (central) and the work done by it is path-independent (conservative), so a gravitational potential energy can be defined.

    Hint: Central = along the join; conservative = path independent.

  5. 5.How does gravitational force between two masses depend on the medium between them?

    It does not depend on the intervening medium. GG is the same everywhere; gravitation cannot be shielded or screened.

    Hint: No screening, unlike electrostatics.

  6. 6.State the principle of superposition for gravitation.

    The net gravitational force on a particle due to several masses is the vector sum of the forces due to each mass taken individually: F=F1+F2+\vec{F}=\vec{F_1}+\vec{F_2}+\dots

    Hint: Vector-add individual pair forces.

  7. 7.How does a uniform spherical shell attract a mass placed outside it?

    It attracts as if its entire mass were concentrated at its centre: F=GMmr2F=\dfrac{GMm}{r^2} with rr measured from the centre.

    Hint: Outside a shell → point mass at centre.

  8. 8.What is the gravitational force on a point mass placed inside a uniform spherical shell?

    Zero. The field inside a uniform shell is zero everywhere, so no net gravitational force acts on a mass inside it.

    Hint: Inside a shell → zero field.

  9. 9.Define acceleration due to gravity gg and give its relation to Earth's mass and radius.

    gg is the acceleration of a freely falling body near Earth's surface: g=GMR2g=\dfrac{GM}{R^2}, where MM and RR are Earth's mass and radius. Value 9.8 m s2\approx 9.8\ \text{m s}^{-2}.

    Hint: g=GM/R2g=GM/R^2.

  10. 10.How is gg related to the mean density ρ\rho of the Earth?

    Using M=43πR3ρM=\dfrac{4}{3}\pi R^3\rho, we get g=43πGRρg=\dfrac{4}{3}\pi G R\rho, so gRρg\propto R\rho for a uniform-density planet.

    Hint: Substitute MM in terms of density.

  11. 11.Distinguish between gg and GG.

    GG is a universal constant (same everywhere, 6.67×10116.67\times10^{-11}). gg is the acceleration due to gravity, a vector that varies with location, altitude, depth and latitude.

    Hint: Universal vs. local; constant vs. variable.

  12. 12.How does gg vary with altitude hh above Earth's surface (general formula)?

    gh=GM(R+h)2=g(RR+h)2g_h=\dfrac{GM}{(R+h)^2}=g\left(\dfrac{R}{R+h}\right)^2. It decreases as height increases.

    Hint: Replace RR by R+hR+h.

  13. 13.Give the approximate expression for the decrease in gg at small altitude hRh\ll R.

    ghg(12hR)g_h\approx g\left(1-\dfrac{2h}{R}\right), so the fractional decrease is Δgg2hR\dfrac{\Delta g}{g}\approx\dfrac{2h}{R}.

    Hint: Binomial approximation, factor 2h/R.

  14. 14.How does gg vary with depth dd below Earth's surface (uniform density)?

    gd=g(1dR)g_d=g\left(1-\dfrac{d}{R}\right). It decreases linearly with depth and becomes zero at the centre.

    Hint: Only the inner sphere of radius R−d matters.

  15. 15.At what point is gg zero, and where is gg maximum?

    g=0g=0 at the centre of the Earth. gg is maximum at the surface (for a uniform-density model) and decreases both above and below the surface.

    Hint: Peak at the surface; zero at centre.

  16. 16.Compare the rate of decrease of gg with altitude versus with depth for small distances.

    For small hh: ghg(12h/R)g_h\approx g(1-2h/R); for small dd: gd=g(1d/R)g_d= g(1-d/R). Thus for equal small distances, gg falls twice as fast going up as going down.

    Hint: Factor 2 for altitude vs 1 for depth.

  17. 17.How does the rotation of the Earth affect gg at latitude λ\lambda?

    gλ=gRω2cos2λg_\lambda=g-R\omega^2\cos^2\lambda, where ω\omega is Earth's angular speed. The rotation reduces the effective gg (except at the poles).

    Hint: Centrifugal term Rω2cos2λR\omega^2\cos^2\lambda.

  18. 18.Where is the effect of Earth's rotation on gg maximum and where is it zero?

    The reduction Rω2cos2λR\omega^2\cos^2\lambda is maximum at the equator (λ=0\lambda=0, cosλ=1\cos\lambda=1) and zero at the poles (λ=90\lambda=90^\circ).

    Hint: cos²λ is 1 at equator, 0 at poles.

  19. 19.Why is gg greater at the poles than at the equator?

    Two reasons: (1) the equatorial radius is larger than the polar radius, and (2) rotation reduces gg at the equator by Rω2R\omega^2. Both make gpole>gequatorg_\text{pole}>g_\text{equator}.

    Hint: Shape (oblate) + rotation.

  20. 20.By roughly what fraction would gg at the equator change if Earth stopped rotating?

    It would increase by Rω2cos2λR\omega^2\cos^2\lambda; at the equator this term is Rω20.034 m s2R\omega^2\approx0.034\ \text{m s}^{-2}, i.e. gg increases by about 0.034 m s20.034\ \text{m s}^{-2}.

    Hint: Add back Rω2R\omega^2 at the equator.

  21. 21.At what angular speed would bodies at the equator become weightless?

    When g=Rω2g=R\omega^2, i.e. ω=g/R1.24×103 rad s1\omega=\sqrt{g/R}\approx1.24\times10^{-3}\ \text{rad s}^{-1}, giving a day of about 8484 minutes.

    Hint: Set effective g to zero at equator.

  22. 22.Define the gravitational field (intensity) at a point.

    It is the gravitational force per unit mass placed at that point: E=Fm\vec{E}=\dfrac{\vec{F}}{m}. Its SI unit is N kg1\text{N kg}^{-1} (same as m s⁻²).

    Hint: Force per unit test mass.

  23. 23.What is the gravitational field due to a point mass MM at distance rr?

    E=GMr2E=\dfrac{GM}{r^2}, directed radially towards the mass. At the surface of Earth this field equals gg.

    Hint: Same form as g = GM/r².

  24. 24.Define gravitational potential VV at a point.

    It is the work done per unit mass in bringing a test mass from infinity to that point: V=WmV=\dfrac{W}{m}. SI unit J kg1\text{J kg}^{-1}; it is always negative (taking zero at infinity).

    Hint: Work per unit mass from infinity.

  25. 25.Write the gravitational potential due to a point mass MM at distance rr.

    V=GMrV=-\dfrac{GM}{r}. It is negative and tends to zero as rr\to\infty.

    Hint: −GM/r; scalar and negative.

Open the interactive deck for the other 72 cards, with self-grading so the ones you keep missing come back.

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