Q.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?

Showing the 12 most recent of 17 on this concept.

1. KCET 2026Set C21 markMCQ
   Q.Imagine a new planet having the same density as that of the earth, but it is two times bigger than the earth in size. If the acceleration due to gravity on the surface of the new planet is g' and that on the surface of the earth is g, then (A) $g' = \frac{g}{4}$ (B) $g' = 8g$ (C) $g' = 2g$ (D) $g' = 4g$

   ›Reveal solutionSolution

   Express surface gravity in terms of density and radius: $g = GM/R^2$ with $M = \tfrac{4}{3}\pi R^3\rho$ gives $g = \tfrac{4}{3}\pi G\rho R$, i.e. $g \propto \rho R$.

   Step 1 — Derive g in terms of density and radius

   $g = \frac{GM}{R^2} = \frac{G\left(\tfrac{4}{3}\pi R^3 \rho\right)}{R^2} = \frac{4}{3}\pi G \rho R$

   Step 2 — Apply the given conditions …

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2. KCET 2026Set C21 markMCQ
   Q.Suppose the acceleration due to gravity at the earth's surface is g m/s$^2$ and at the surface of moon it is g' m/s$^2$. An M kg passenger goes from the earth to the moon in a spaceship moving with a constant velocity (Neglect all other objects in the sky). Which curve best represents the weight (net gravitational force) as a function of time? [FIGURE: A graph of weight (y-axis, marked Mg at top and Mg' lower down) versus time t (x-axis, up to t_s), showing four labelled curves A, B, C, D all starting at Mg; curves A and B decline toward the right without reaching zero; curve C dips to near zero at a point between O and t_s then rises back up to a value near Mg' at t_s; curve D dips lowest (to zero) between O and t_s and then rises back up, ending at Mg' at t_s.] (A) A (B) B (C) C (D) D

   ›Reveal solutionSolution

   The net gravitational force (weight) on the passenger is the vector sum of Earth's pull (toward Earth, dominant near $t=0$) and the Moon's pull (toward the Moon, dominant near $t_s$); between them lies a neutral point where these two pulls exactly cancel.

   Step 1 — Write the net force as a function of position

   At a distance $r$ from Earth along a line of length $D$ to the Moon, the net gravitational force on the passenger of mass $M$ is

   $F_{net} = \frac{GM_e M}{r^2}\ (\text{toward Earth}) \;-\; \frac{GM_{moon} M}{(D-r)^2}\ (\text{toward Moon})$

   At $t=0$ (near Earth), $F_{net} \approx Mg$, matching every curve's starting point.

   Step 2 — Locate the neutral point

   As $r$ increases, Earth's pull weakens as $1/r^2$ while the Moon's pull strengthens as $1/(D-r)^2$. Since $M_{moon} \ll M_e$, this neutral point (where the two forces are exactly equal and opposite, giving $F_{net}=0$) lies quite close to the Moon, but it is reached well before $t_s$.

   Step 3 — Behaviour after the neutral point …

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3. KCET 2025Set D-41 markMCQ
   Q.The total energy of a satellite in a circular orbit at a distance $(R + h)$ from the centre of the Earth varies as [R is the radius of the Earth and h is the height of the orbit from Earth’s surface] (A) $-\dfrac{1}{(R + h)}$ (B) $-\dfrac{1}{(R + h)^2}$ (C) $+\dfrac{1}{(R + h)^2}$ (D) $+\dfrac{1}{(R + h)}$

   ›Reveal solutionSolution

   The total energy of a satellite in a circular orbit is the sum of its kinetic and potential energies, which together give a dependence of $-\frac{1}{(R+h)}$, so the correct option is (A).

   The key idea is that for any satellite in a stable circular orbit, the gravitational force provides the necessary centripetal acceleration. This links the orbital speed directly to the orbital radius, and from there the total mechanical energy (kinetic + potential) simplifies to a neat expression.

   Let’s build this from the ground up.

   1. Gravitational potential energy For a satellite of mass $m$ at a distance $r = R + h$ from the Earth’s centre, the gravitational potential energy is

   $U = -\frac{GMm}{r}$

   where $M$ is the Earth’s mass. This is always negative because gravity is attractive — zero only at infinite separation.

   2. Kinetic energy from orbital motion In a circular orbit, the centripetal force is supplied entirely by gravity:

   $\frac{GMm}{r^2} = \frac{mv^2}{r}$

   Cancel $m$ and multiply through by $r$:

   $v^2 = \frac{GM}{r}$

   So the kinetic energy is

   $K = \frac{1}{2}mv^2 = \frac{1}{2}m \cdot \frac{GM}{r} = \frac{GMm}{2r}$

   3. Total mechanical energy Add potential and kinetic:

   $E = K + U = \frac{GMm}{2r} - \frac{GMm}{r} = -\frac{GMm}{2r}$

   This is the famous result: for a circular orbit, the total energy is exactly half the potential energy (and negative). …

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4. KCET 2025Set D-41 markMCQ
   Q.Which of the following statements is incorrect with reference of ‘Nuclear force’? (A) Nuclear force becomes attractive for nucleon distances larger than $0.8\,\mathrm{fm}$ (B) Nuclear force becomes repulsive for nucleon distances less than $0.8\,\mathrm{fm}$ (C) Nuclear force is always attractive (D) Potential energy is minimum, if the separation between the nucleons is $0.8\,\mathrm{fm}$

   ›Reveal solutionSolution

   Read the nucleon–nucleon potential-energy curve: it has a minimum at $r_0 \approx 0.8\,\mathrm{fm}$, so the force is attractive outside that distance and repulsive inside it — hence "always attractive" is the false statement.

   Step 1 — The shape of the nuclear potential.

   The potential energy $U(r)$ of two nucleons falls steeply from large positive values at very small $r$, passes through a minimum at $r_0 \approx 0.8\,\mathrm{fm}$, and then rises towards zero as $r$ grows, becoming negligible beyond a few femtometres.

   Step 2 — Force from the potential.

   The force is the negative gradient of the potential:

   $F(r) = -\frac{dU}{dr}$

   • For $r > 0.8\,\mathrm{fm}$: $U$ is increasing with $r$, so $dU/dr > 0$ and $F < 0$ — the force pulls the nucleons together, i.e. attractive. This is what binds the nucleus.
   • For $r < 0.8\,\mathrm{fm}$: $U$ is decreasing with $r$, so $dU/dr < 0$ and $F > 0$ — the force pushes the nucleons apart, i.e. repulsive. This short-range repulsive "hard core" is why nuclear matter has a nearly constant density and does not collapse.
   • At $r = 0.8\,\mathrm{fm}$: $dU/dr = 0$, so $F = 0$ — the equilibrium separation, where $U$ is minimum.

   Step 3 — Test each option. …

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5. COMEDK 2025Set 2025-A1 markMCQ
   Q.If the radius of earth were to shrink by two percent, its mass remaining the same, the acceleration due to gravity on the earth's surface would (A) Decrease by $4 \%$ (B) Increase by $1 \%$ (C) Increase by $4 \%$ (D) Increase by $2 \%$

   ›Reveal solutionSolution

   The acceleration due to gravity on Earth’s surface depends inversely on the square of the radius. If the radius shrinks by 2% (mass constant), gravity increases by about 4%. The correct option is (C).

   Concept & Intuition

   The acceleration due to gravity at the surface of a planet is given by

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

   where $G$ is the gravitational constant, $M$ is the planet’s mass, and $R$ is its radius.

   If the mass stays the same but the radius decreases, the surface gets closer to the center of mass. Since gravity follows an inverse-square law, a small reduction in radius causes a larger relative increase in $g$. A 2% decrease in $R$ means the new radius is $0.98R$, so $g$ becomes $\frac{GM}{(0.98R)^2} = \frac{GM}{0.9604 R^2} \approx 1.041\,g$. That’s roughly a 4% increase.

   Tip

   For small changes, we can use the approximation:

   If $R$ changes by a small fraction $x$ (here $x = -0.02$), then $g$ changes by approximately $-2x$. Since $x$ is negative, $-2x$ is positive, giving about a 4% increase. This avoids squaring decimals.

   Now let’s work it through step by step.

   1. Write the formula for $g$

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

   Here $G$ and $M$ are constants, so $g$ is inversely proportional to $R^2$.

   2. Express the new radius A 2% shrinkage means the new radius $R'$ is

   $R' = R - 0.02R = 0.98R.$

   3. Find the new acceleration $g'$

   $g' = \frac{GM}{(R')^2} = \frac{GM}{(0.98R)^2} = \frac{GM}{0.9604\,R^2}.$

   Since $\frac{GM}{R^2} = g$, we have

   $$$$

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6. COMEDK 2025Set 2025-M1 markMCQ
   Q.The radius of earth is $R$ and acceleration due to gravity on its surface is $g$. The height at which the acceleration due to gravity becomes $\frac{g}{8}$ is: (A) $2 R$ (B) $(2 \sqrt{2}-1) R$ (C) $\sqrt{2} R$ (D) $2 \sqrt{2} R$

   ›Reveal solutionSolution

   The acceleration due to gravity decreases with height according to an inverse-square law. At height $h$, $g_h = g \frac{R^2}{(R+h)^2}$. Setting $g_h = g/8$ gives $h = (\sqrt{8} - 1)R = (2\sqrt{2} - 1)R$, so the correct option is (B).

   The key idea is that gravity follows an inverse-square law with distance from the Earth’s center. On the surface, distance is $R$; at height $h$, distance is $R+h$. The acceleration drops as the square of the distance increases. So to get $g/8$, the distance must be multiplied by $\sqrt{8} = 2\sqrt{2}$. That directly gives the height.

   Let’s work it through:

   1. Write the formula for gravity at height $h$ The acceleration due to gravity at a distance $r$ from Earth’s center is

   $g(r) = \frac{GM}{r^2}$

   On the surface, $r = R$ and $g(R) = g = \frac{GM}{R^2}$.

   At height $h$ above the surface, $r = R + h$, so

   $g_h = \frac{GM}{(R+h)^2} = g \cdot \frac{R^2}{(R+h)^2}.$

   2. Set the condition We want $g_h = \frac{g}{8}$. Substitute:

   $g \cdot \frac{R^2}{(R+h)^2} = \frac{g}{8}.$

   Cancel $g$ (non-zero):

   $\frac{R^2}{(R+h)^2} = \frac{1}{8}.$

   3. Solve for $R+h$ Take reciprocals:

   $\frac{(R+h)^2}{R^2} = 8 \quad \Rightarrow \quad \left( \frac{R+h}{R} \right)^2 = 8.$

   Take the positive square root (distance is positive):

   $\frac{R+h}{R} = \sqrt{8} = 2\sqrt{2}.$

   So

   $$$$

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7. COMEDK 2024Set 2024-A1 markMCQ
   Q.A planet has double the mass of the earth and double the radius. The gravitational potential at the surface of the Earth is $\mathrm{V}$ and the magnitude of the gravitational field strength is $\mathrm{g}$. The gravitational potential and gravitational field strength on the surface of the planet are .tg {border-collapse:collapse;border-spacing:0;} .tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px; overflow:hidden;padding:10px 5px;word-break:normal;} .tg th{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px; font-weight:normal;overflow:hidden;padding:10px 5px;word-break:normal;} .tg .tg-baqh{text-align:center;vertical-align:top} .tg .tg-amwm{font-weight:bold;text-align:center;vertical-align:top} Potential Field A V $\frac{g}{4}$ B 2V $\frac{g}{2}$ C V $\frac{g}{2}$ D 2V $\frac{g}{4}$ (A) C (B) D (C) A (D) B

   ›Reveal solutionSolution

   Doubling both mass and radius leaves the surface potential unchanged ($V$) and halves the field strength ($g/2$) — that is table row C, i.e. option (A).

   Earth: potential $V=-\dfrac{GM}{R}$, field magnitude $g=\dfrac{GM}{R^{2}}$.

   Planet: $M'=2M$, $R'=2R$.

   Potential:

   $V'=-\frac{GM'}{R'}=-\frac{G(2M)}{2R}=-\frac{GM}{R}=V.$

   Field strength:

   $g'=\frac{GM'}{R'^{2}}=\frac{G(2M)}{(2R)^{2}}=\frac{2GM}{4R^{2}}=\frac{g}{2}.$ …

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8. COMEDK 2024Set 2024-E1 markMCQ
   Q.The acceleration due to gravity at pole and equator can be related as (A) $g_e=g_p < g$ (B) $g_e=g_p=g$ (C) $g_e < g_p$ (D) $g_e > g_p$

   ›Reveal solutionSolution

   The acceleration due to gravity is slightly less at the equator than at the poles because of Earth’s rotation and its equatorial bulge. Thus $g_e < g_p$, making option (C) correct.

   Why this is the right approach

   The value of $g$ (acceleration due to gravity) is not constant over Earth’s surface. Two main effects cause it to vary:

   1. Earth’s rotation – At the equator, part of the gravitational force is “used up” to keep objects moving in a circle (centripetal force), so the measured weight is slightly less.
   2. Earth’s shape – Earth is an oblate spheroid: the equatorial radius is larger than the polar radius. Since $g \propto 1/R^2$, gravity is weaker at the equator.

   Both effects work in the same direction: they reduce $g$ at the equator relative to the poles.

   ---

   Step-by-step reasoning

   1. Recall the formula for apparent gravity The effective acceleration due to gravity at a latitude $\phi$ is

   $g_{\text{eff}} = g_0 - \omega^2 R \cos^2\phi$

   where $g_0$ is the gravity if Earth were non-rotating, $\omega$ is Earth’s angular speed, and $R$ is the local radius. At the pole ($\phi = 90^\circ$), $\cos\phi = 0$, so

   $g_p = g_0$

   At the equator ($\phi = 0^\circ$), $\cos\phi = 1$, so

   $g_e = g_0 - \omega^2 R$

   Since $\omega^2 R > 0$, we get $g_e < g_p$.

   2. Consider the shape effect Earth’s equatorial radius is about 21 km larger than the polar radius. Using the inverse-square law:

   $g \propto \frac{1}{R^2}$

   A larger $R$ at the equator means a smaller $g$. This reinforces the inequality $g_e < g_p$. …

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9. KCET 2023Set A-31 markMCQ
   Q.A closed water tank has cross-sectional area $A$. It has a small hole at a depth $h$ from the free surface of water. The radius of the hole is $r$ so that $r \ll \sqrt{\dfrac{A}{\pi}}$. If $P_0$ is the pressure inside the tank above water level, and $P_a$ is the atmospheric pressure, the rate of flow of the water coming out of the hole is [$\rho$ is the density of water] (A) $\pi r^2 \sqrt{2gh + \dfrac{2(P_0 - P_a)}{\rho}}$ (B) $\pi r^2 \sqrt{2gH}$ (C) $\pi r^2 \sqrt{gh + \dfrac{2(P_0 - P_a)}{\rho}}$ (D) $\pi r^2 \sqrt{2gh}$

   ›Reveal solutionSolution

   The key idea is to apply Bernoulli’s equation between the free surface and the hole, accounting for the excess pressure $P_0$ above the water. The efflux speed is $v = \sqrt{2gh + \frac{2(P_0 - P_a)}{\rho}}$, and the flow rate is area times speed, giving option (A).

   The problem is a classic extension of Torricelli’s theorem. When the tank is open to the atmosphere, the efflux speed is simply $\sqrt{2gh}$. Here, the tank is closed and the air above the water is at a pressure $P_0$ that may differ from atmospheric pressure $P_a$ outside the hole. That pressure difference adds an extra term to the kinetic energy of the emerging water.

   The condition $r \ll \sqrt{A/\pi}$ means the hole area is tiny compared to the tank’s cross-section, so the water level falls very slowly. This lets us treat the flow as steady and the speed of the free surface as negligible.

   1. Choose two points for Bernoulli’s equation.

      Point 1: at the free surface inside the tank, where pressure is $P_0$, speed is nearly $0$, and height is taken as $0$ (reference level).

      Point 2: just outside the hole, where pressure is $P_a$ (atmospheric), speed is $v$ (what we want), and height is $-h$ (since the hole is $h$ below the surface).

   2. Write Bernoulli’s equation for an ideal, incompressible fluid:

   $P_0 + \frac{1}{2}\rho (0)^2 + \rho g(0) = P_a + \frac{1}{2}\rho v^2 + \rho g(-h)$

   This simplifies to:

   $P_0 = P_a + \frac{1}{2}\rho v^2 - \rho g h$

   3. Solve for $v$:

   $\frac{1}{2}\rho v^2 = P_0 - P_a + \rho g h$

   $v^2 = \frac{2(P_0 - P_a)}{\rho} + 2gh$

   $v = \sqrt{2gh + \frac{2(P_0 - P_a)}{\rho}}$ …

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10. COMEDK 2023Set 2023-E1 markMCQ
    Q.The acceleration due to gravity at a height of $7 \mathrm{~km}$ above the earth is the same as at a depth d below the surface of the earth. Then d is (A) 7 km (B) 2 km (C) 3.5 km (D) 14 km

    ›Reveal solutionSolution

    Since $g$ falls twice as fast with height as with depth ($g_h=g(1-2h/R)$ vs $g_d=g(1-d/R)$), equal values require $d=2h=14$ km.

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

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

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11. COMEDK 2023Set 2023-M1 markMCQ
    Q.Starting from the centre of the earth having radius $R$, the variation of $g$ (acceleration due to gravity) is shown by (A) (B) (C) (D)

    ›Reveal solutionSolution

    The acceleration due to gravity inside a uniform sphere increases linearly with distance from the centre, then outside it falls off as $1/r^2$. The correct graph is a straight-line rise from zero at the centre to a maximum at the surface, followed by a curved decay — this matches option (B).

    The key concept is Newton’s shell theorem and how gravity behaves inside and outside a uniform sphere.

    • Inside a uniform sphere ($r < R$), only the mass inside radius $r$ contributes to gravity; that mass is proportional to $r^3$, so $g \propto r$.
    • Outside ($r > R$), the entire mass acts as if concentrated at the centre, so $g \propto 1/r^2$.

    Thus the graph must show a linear increase from $g=0$ at the centre to a maximum at the surface, then a smooth, concave-up decay (not a straight line) for $r > R$. Let’s check each option.

    1. Inside the Earth ($0 \le r \le R$) For a uniform sphere of radius $R$ and mass $M$, the mass enclosed at radius $r$ is

    $M_{\text{enc}} = \frac{M}{R^3} r^3.$

    By Newton’s law,

    $g(r) = \frac{G M_{\text{enc}}}{r^2} = \frac{G M}{R^3} r.$

    So $g$ is directly proportional to $r$ — a straight line from the origin. This eliminates option (C) (curved rise) and option (D) (starts at maximum, no rise).

    2. At the surface ($r = R$)

    $g(R) = \frac{G M}{R^2},$

    the familiar surface gravity. The graph must peak exactly at $r = R$.

    3. Outside the Earth ($r > R$) The entire mass $M$ acts at the centre, so $g(r) = \frac{G M}{r^2}.$ …

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12. COMEDK 2023Set 2023-M1 markMCQ
    Q.The height vertically above the earth's surface at which the acceleration due to gravity becomes $1 \%$ of its value at the surface is (A) 8R (B) 9R (C) 10R (D) 20R

    ›Reveal solutionSolution

    Setting $g(R/(R+h))^2 = 0.01\,g$ gives $R/(R+h)=0.1$, hence $R+h=10R$ and $h=9R$.

    Above the surface,

    $g' = g\left(\frac{R}{R+h}\right)^2.$

    We need $g'=0.01\,g$: …

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