Q.Find the acceleration due to gravity at a height above the Earth's surface equal to the Earth's own radius $R$. Take $g = 9.8\ \text{m/s}^2$ at the surface.

Showing the 12 most recent of 22 on this concept.

1. CBSE 2026Set ANNUAL1 mark
   Q.The value of an acceleration of a freely falling body is ______.

   ›Reveal solutionSolution

   A freely falling body (falling under gravity alone, no air resistance) accelerates at the constant value g ≈ 9.8 m/s^2 downward.

   Near the Earth's surface, every freely falling object experiences the same acceleration g, independent of its mass, given approximately by g = GM/R^2 where G is the universal gravitational constant, M is Earth's mass and R is Earth's radius. Its standard value is g ≈ 9.8 m/s^2 (often rounded to 9.81 m/s^2 or taken as 10 m/s^2 for quick calculations), always directed vertically downward.

   ✓Final answer

   g ≈ 9.8 m/s^2, directed vertically downward.

2. CBSE 2026Set ANNUAL1 markMCQ
   Q.Correct relation between g and density (ρ) of earth is:

   (a) g = (3/4)Rρ

   (b) g = (4/3)πRρ

   (c) g = (4/3)πGRρ

   (d) g = (3/4)πGRρ

   ›Reveal solutionSolution

   $g=GM/R^2$ combined with $M=\tfrac43\pi R^3\rho$ (sphere of density $\rho$) gives $g=\tfrac43\pi GR\rho$.

   The acceleration due to gravity at the Earth's surface is:

   $g = \frac{GM}{R^2}$

   Modeling the Earth as a uniform sphere of radius $R$ and density $\rho$, its mass is:

   $M = \rho\times\text{Volume} = \rho\times\frac43\pi R^3$

   Substituting:

   $g = \frac{G}{R^2}\times\frac43\pi R^3\rho = \frac43\pi G R\rho$

   ✓Final answer

   (c) $g = \dfrac43\pi G R\rho$.

3. CBSE 2026Set ANN1 mark
   Q.At the centre of Earth, the value of g is ________ .

   ›Reveal solutionSolution

   At the Earth's centre the acceleration due to gravity is zero, because no mass is enclosed below that point to produce a net gravitational pull.

   Inside a uniform sphere, only the mass contained within the radius r at which the body sits contributes to the gravitational field (the outer shells exert no net force on a point inside them). The value of g at depth d below the surface is

   g_d = g (1 - d/R),

   where R is the Earth's radius. At the centre the depth d = R, so

   g_centre = g (1 - R/R) = g x 0 = 0.

   Physically, at the centre the mass is distributed symmetrically all around, and the pulls in all directions cancel, giving zero net gravity.

   ✓Final answer

   Zero.

4. CBSE 2025Set ANNUAL1 markMCQ
   Q.Acceleration due to gravity for the earth is maximum at (A) equator (B) pole (C) centre (D) none of these

   ›Reveal solutionSolution

   Acceleration due to gravity is maximum at the poles and minimum at the equator.

   Two effects reduce $g$ as one moves from the pole to the equator:

   1. Earth's shape: earth is an oblate spheroid, slightly flattened at the poles and bulging at the equator, so the equatorial radius is larger than the polar radius. Since $g \propto 1/R^2$, a larger radius at the equator means smaller $g$.
   2. Earth's rotation: the effective gravity felt is reduced by the centrifugal effect of rotation, $g_{eff} = g - \omega^2R\cos^2\lambda$, which is largest (most reduction) at the equator ($\lambda=0$) and zero at the poles ($\lambda=90^\circ$).

   Both effects make $g$ largest at the poles. (At the centre, $g=0$, since it lies inside the earth with equal pull from all sides.)

   ✓Final answer

   (B) pole.

5. CBSE 2025Set ANNUAL1 markMCQ
   Q.A body weighs 200 N on the surface of the earth. How much will it weigh half way down to the center of the earth?

   (a) 100 N

   (b) 50 N

   (c) 200 N

   (d) 400 N

   ›Reveal solutionSolution

   Unlike the inverse-square fall-off above the surface, gravity inside a uniform earth decreases linearly with depth — so at half the radius, both $g$ and the weight are exactly halved.

   Assuming the earth is a uniform sphere of radius $R$, the acceleration due to gravity at depth $d$ below the surface is:

   $g_d = g\left(1 - d/R\right)$

   'Half way down to the centre' means $d = R/2$:

   $g_d = g\left(1 - (R/2)/R\right) = g(1 - 1/2) = g/2$

   Weight at that depth:

   $W_d = m g_d = m(g/2) = W_{surface}/2 = 200/2 = 100$ N

   ✓Final answer

   Weight halfway down to the centre $= 100$ N — option (a).

6. CBSE 2025Set sz1 markMCQ
   Q.At what depth below the surface of earth the value of 'g' is same as that of height of 5 km? (A) 10 km (B) 7.5 km (C) 5 km (D) 2.5 km

   ›Reveal solutionSolution

   Setting the depth expression equal to the height expression gives d = 2h = 10 km.

   At height h above the surface: $g_h = g\left(1 - \dfrac{2h}{R}\right)$.

   At depth d below the surface: $g_d = g\left(1 - \dfrac{d}{R}\right)$.

   For the two to be equal: $1 - \dfrac{d}{R} = 1 - \dfrac{2h}{R} \Rightarrow d = 2h$.

   With h = 5 km, $d = 2 \times 5 = 10$ km.

   ✓Final answer

   The correct option is (A) 10 km.

7. CBSE 2024Set ANNUAL1 markMCQ
   Q.If the earth starts rotating from east to west instead of west to east about its axis, then the value of g on equator (A) will increase (B) will decrease (C) will be same (D) will be zero

   ›Reveal solutionSolution

   Reversing earth's spin direction doesn't change $g$ at the equator, since the correction term depends on $\omega^2$.

   At the equator, the effective (measured) value of gravity is reduced from the true gravitational value by the centrifugal effect of the earth's rotation: $g_{eq}=g_{true}-\omega^2R$, where $\omega$ is the earth's angular speed and $R$ its equatorial radius.

   Reversing the direction of rotation (east-to-west instead of west-to-east) changes the sign of $\omega$ as a vector, but $\omega^2$ (which is what appears in the centrifugal term) stays exactly the same. So $g_{eq}$ is unaffected.

   ✓Final answer

   (C) will be same.

8. CBSE 2024Set ANNUAL1 markMCQ
   Q.Where will it be profitable to purchase one kilogram of sugar?

   (a) At poles

   (b) At equator

   (c) At 45 degree latitude

   (d) At 40 degree latitude

   ›Reveal solutionSolution

   Because g is minimum at the equator (due to Earth's bulge and rotation), buying sugar 'by weight' there gets you more actual mass for the same weight reading — so it is profitable to buy at the equator.

   A spring balance measures WEIGHT (W = mg), not mass directly. Earth's value of g is slightly larger at the poles and slightly smaller at the equator (due to Earth's equatorial bulge and the effect of its rotation reducing effective gravity at the equator).

   If you buy '1 kg' of sugar going by a spring-balance weight reading, you're really buying an amount whose weight (mg) matches a fixed reading. Since g is smaller at the equator, a LARGER mass m is needed to produce the same weight reading there compared to at the poles. So at the equator, a '1 kg-weight' worth of sugar actually corresponds to a slightly greater mass of sugar — making it more profitable to purchase there.

   ✓Final answer

   (b) At equator.

9. CBSE 2024Set SET-AP55001 markMCQ
   Q.At height h above the earth's surface, the value of g changes by the same amount as it does at depth x inside the earth (both x and h are much smaller than the earth's radius) when:

   (a) x = h

   (b) x = h/2

   (c) x = 2h

   (d) x = h^2

   ›Reveal solutionSolution

   For small h, R: g at height h is g(1 − 2h/R); g at depth x is g(1 − x/R). Equating the fractional drops gives x = 2h.

   At height h above Earth's surface (h << R):

   g_h = g / (1 + h/R)^2 ≈ g(1 − 2h/R)

   so the fractional decrease is Δg_h/g ≈ 2h/R.

   At depth x below Earth's surface (x << R):

   g_x = g(1 − x/R)

   so the fractional decrease is Δg_x/g = x/R.

   For the two changes to be equal:

   2h/R = x/R

   ⟹ x = 2h

   This is a standard result: going a certain height up decreases g about twice as fast (fractionally) as going the same distance down, so you need to go down twice as far to match the same drop.

   ✓Final answer

   The correct option is (c) x = 2h.

10. CBSE 2024Set SET-NDP60001 markMCQ
    Q.The value of 'g' is maximum at:

    (a) At surface of earth

    (b) At height

    (c) At depth

    (d) At center

    ›Reveal solutionSolution

    g is maximum at the earth's surface and decreases both with height above it and with depth below it.

    Above the surface, at height $h$, $g_h = g\left(1 - \dfrac{2h}{R}\right)$ (for $h \ll R$) — this is always less than the surface value $g$. Below the surface, at depth $d$, $g_d = g\left(1 - \dfrac{d}{R}\right)$ — this too is always less than $g$, and $g$ falls all the way to zero at the centre of the earth ($d = R$). So moving away from the surface in either direction (up or down) reduces $g$; the surface itself is where $g$ takes its maximum value (for a uniform-density spherical earth, ignoring the small effect of the earth's rotation and non-uniform density in reality).

    ✓Final answer

    The correct option is (a) At surface of earth.

11. CBSE 2024Set ANNUAL1 markMCQ
    Q.Value of acceleration due to gravity with increasing depth from earth surface :

    (a) increases

    (b) decreases

    (c) remains unaltered

    (d) None

    ›Reveal solutionSolution

    Below the earth's surface, only the mass enclosed within radius (R−d) contributes to gravity, so g falls linearly with depth.

    At depth d below the earth's surface, treating the earth as a uniform sphere of radius R and density ρ, only the mass of the sphere of radius (R − d) contributes to the gravitational field at that depth (the shell outside contributes zero net field). This gives:

    $g_d = g\left(1 - \frac{d}{R}\right)$

    As d increases from 0 (surface) toward R (centre), the factor $(1 - d/R)$ decreases from 1 toward 0. So g decreases with depth, becoming zero at the centre of the earth.

    ✓Final answer

    Acceleration due to gravity decreases with increasing depth below the earth's surface. Option (b) decreases.

12. CBSE 2024Set ANNUAL1 markMCQ
    Q.Match the values of Physical quantities given in Column I with the Numerical values given in Column II: Column-I (I) Acceleration due to gravity at the centre of earth (II) Escape speed of earth (III) Universal Gravitational constant (IV) Acceleration due to gravity Column-II (A) 9.8 ms^-2 (B) 6.67 x 10^-11 Nm^2 kg^-2 (C) 0 (D) 11.2 kms^-1

    (a) (I)-D (II)-A (III)-B (IV)-C

    (b) (I)-C (II)-A (III)-D (IV)-B

    (c) (I)-A (II)-B (III)-D (IV)-C

    (d) (I)-C (II)-D (III)-B (IV)-A

    ›Reveal solutionSolution

    (I) g at earth's centre = 0 (C); (II) escape speed = 11.2 km/s (D); (III) G = 6.67x10^-11 N m^2 kg^-2 (B); (IV) g at surface = 9.8 m/s^2 (A). This is option (d).

    Matching each physical quantity to its numerical value:

    (I) Acceleration due to gravity at the centre of the earth: treating the earth as a uniform sphere, the gravitational field inside a uniform spherical shell is zero everywhere inside it (shell theorem), so at the very centre, contributions from all directions cancel and g = 0 -> matches (C).

    (II) Escape speed of earth: v_e = sqrt(2GM/R) ~ 11.2 km/s, the minimum speed needed for an object to escape earth's gravitational field from the surface without further propulsion -> matches (D).

    (III) Universal Gravitational constant G = 6.67 x 10^-11 N m^2 kg^-2, the constant of proportionality in Newton's law of gravitation F = G m1 m2/r^2 -> matches (B).

    (IV) Acceleration due to gravity (at the earth's surface) ~ 9.8 m s^-2, the standard value used for g near the earth's surface -> matches (A).

    So the correct matching is (I)-C, (II)-D, (III)-B, (IV)-A.

    ✓Final answer

    The correct option is (d) (I)-C (II)-D (III)-B (IV)-A.