Q.If the mass of sun were ten times smaller and gravitational constant G were ten times larger in magnitudes- (Note: more than one of the given options may be correct.)
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🔒 Start your 14-day free trial to unlock the full solution →Concept understanding — Variation Of Gravity
Variation of Gravity: Why Your Weight Changes Even When You Don't
Imagine you step on a weighing scale at sea level in Mumbai, then carry that same scale to the top of Mount Everest. The scale would show a smaller number — you'd weigh less. But you haven't lost any mass. What changed?
The force pulling you down — gravity — is not constant everywhere on Earth. It varies. That's what we mean by variation of gravity.
The Core Idea
Gravity is the force with which the Earth pulls objects toward its centre. The strength of this pull depends on two things: the mass of the Earth and your distance from its centre. Since the Earth is not a perfect sphere and it spins, that distance and the effective pull change from place to place.
The acceleration due to gravity, denoted by g, is approximately 9.8m/s2 at sea level. But that's an average. The actual value can be slightly higher or lower depending on where you are.
Why Does Gravity Vary? Three Main Reasons
1. Altitude (Height Above Sea Level)
This is the most intuitive one. As you go higher, you move farther from the Earth's centre. Gravity follows an inverse-square law: double the distance, and the force becomes one-fourth.
The formula for g at a height h above the Earth's surface (where R is Earth's radius, about 6400 km) is:
gh=(R+h)2GM
For small heights compared to R, we can approximate:
gh≈g(1−R2h)
This means for every kilometre you go up, g decreases by roughly 0.003m/s2. That's why at the top of a tall mountain, you weigh about 0.5% less than at sea level.
2. Depth (Going Underground)
What happens if you go down a mine or into the Earth's crust? Intuition might say gravity increases because you're closer to the centre. But the opposite happens.
Inside the Earth, the mass above you pulls upward, partially cancelling the pull from below. For a uniform Earth, only the mass inside the sphere of radius r (your distance from the centre) contributes to gravity at that point.
gd=r2GM′
Where M′ is the mass of the sphere of radius r. If Earth had uniform density ρ, then M′=34πr3ρ, giving:
gd=34πGρr
This means gravity decreases linearly as you go deeper. At the centre of the Earth, g=0 — you'd be weightless, pulled equally in all directions.
This linear decrease assumes uniform density. The real Earth has a dense iron core, so the actual variation is more complicated — gravity actually increases slightly as you go down through the crust before eventually decreasing.
3. Rotation of the Earth (Latitude Effect)
The Earth spins once every 24 hours. This rotation creates a centrifugal force that acts outward, away from the axis of rotation. This force effectively reduces the weight you feel.
The effect is strongest at the equator (where the rotational speed is highest, about 1670 km/h) and zero at the poles (where you're on the axis of rotation).
The effective g at latitude ϕ is:
geff=g−ω2Rcos2ϕ
Where ω is Earth's angular speed (7.3×10−5rad/s) and R is Earth's radius.
At the equator (ϕ=0∘), the reduction is about 0.034m/s2 — roughly 0.35% of g.
| Location | Approximate g (m/s²) | Why? |
|----------|------------------------|------|
| Equator (sea level) | 9.78 | Fastest rotation + bulging equator |
| 45° latitude | 9.81 | Intermediate |
| North Pole | 9.83 | No rotation effect + closer to centre |
4. Shape of the Earth (Oblateness)
The Earth is not a perfect sphere. Because of its rotation, it bulges at the equator and flattens at the poles. The equatorial radius is about 21 km larger than the polar radius.
This means:
- At the poles, you're closer to the Earth's centre → stronger gravity
- At the equator, you're farther from the centre → weaker gravity …
The key idea is variation of gravity: the acceleration due to gravity on Earth depends on Earth's mass Me and radius Re as g=Re2GMe, while the Sun's mass affects orbital motion and solar tides, not g directly.
- Walking difficulty depends on g. With G ten times larger and Sun's mass ten times smaller, Earth's mass Me is unchanged, so g becomes 10 times larger (since g∝G). This makes walking harder — option (A) is correct.
- Since g changes, option (B) is false. …
Earth's surface gravity g=GM_Earth/R_Earth^2 depends on G and Earth's own mass/radius, not the Sun's mass. With G becoming ten times larger, g becomes ten times larger -- making walking harder, raindrops fall faster, and airplanes need to fly faster. Correct options: (A), (C), (D).
- Surface gravity depends on G and Earth's own mass, not the Sun's mass. g=GM_Earth/R_Earth^2. With G to 10G, M_Earth and R_Earth unchanged, g to 10g. Option (B) is false.
- Walking becomes harder -- option (A). Weight mg becomes ten times larger, so far more muscular effort and friction are needed.
- Raindrops fall faster -- option (C). Terminal velocity v_t is proportional to sqrt(g); a ten-fold g increase raises v_t by about 3.16x. …
Step 1: surface gravity g=GM_Earth/R_Earth^2 depends on G and Earth's own mass/radius, not the Sun's mass. With G->10G, M_Earth and R_Earth unchanged, g->10g. Step 2: weight W=mg becomes ten times larger -- walking requires far more muscular effort -- confirms (a); (b) false. Step 3: terminal velocity of raindrops scales as sqrt(g) -- a ten-fold g increase raises v_t by ~3.16x -- confirms (c). Step 4: airplane lift must balance a ten-times-heavier weight; since lift is proportional to v^2, required speed scales as sqrt(g) -- …
- KEAM 2026Set eng-2026-04174 marksMCQQ.The height above the surface of the earth at which the acceleration due to gravity becomes 9g in terms of radius of earth R is (g is acceleration due to gravity at the surface of the earth) (A) 4R (B) 3R (C) 2R (D) 2R (E) 3R
›Reveal solutionSolution
Setting g/(1+h/R)2=g/9 gives h=2R.
The acceleration due to gravity at a height h above the earth's surface is:
g′=(1+Rh)2g
Setting g′=g/9: …
- KEAM 2026Set eng-2026-04184 marksMCQQ.The values of the acceleration due to gravity at depths of 2R and 32R from the surface of the earth are in the ratio (A) 9 : 10 (B) 3 : 2 (C) 3 : 1 (D) 1 : 3 (E) 2 : 1
›Reveal solutionSolution
With gd=g(1−Rd), the depths R/2 and 2R/3 give g/2 and g/3, a ratio 3:2.
The value of gravity at depth d is gd=g(1−Rd).
At d=2R: gd=g(1−21)=2g. …
- KEAM 2026Set eng-2026-04214 marksMCQQ.The ratio of the weights of an object of mass m at a height R and 2R from the surface of earth is (R is the radius of earth) (A) 4 : 9 (B) 1 : 1 (C) 9 : 4 (D) 1 : 2 (E) 4 : 1
›Reveal solutionSolution
Using gh=gR2/(R+h)2, the weights at R and 2R are in ratio 9:4.
Acceleration due to gravity at height h: gh=g(R+h)2R2, and weight =mgh.
At h=R: gR=g(2R)2R2=4g. …
- KEAM 2026Set eng-2026-04224 marksMCQQ.The height at which the weight of a body becomes one third of its weight on the surface of earth is (R is the radius of earth) (A) 3R (B) 0.732R (C) 0.414R (D) 6R (E) 10R
›Reveal solutionSolution
Weight varies as g∝1/(R+h)2; set it to one third of surface value and solve for h.
ggh=(R+hR)2=31. …
- KEAM 2025Set eng-2025-04234 marksMCQQ.There is a mine of depth about 3.0 km. Conditions prevailing in this mine as compared to those at the surface of earth are (A) higher air pressure, lower acceleration due to gravity (B) higher air pressure, higher acceleration due to gravity (C) lower air pressure, higher acceleration due to gravity (D) lower air pressure, lower acceleration due to gravity (E) same air pressure and acceleration due to gravity
›Reveal solutionSolution
In a deep mine the air pressure is higher and the acceleration due to gravity is lower than at the surface.
Concept and Intuition
Going down into a mine, there is more atmosphere above you, so air pressure increases. At the same time g decreases below the Earth's surface as gd=g(1−d/R), because only the mass of the sphere below your depth contributes.
Step-by-Step Solution
- More air column above ⇒ higher atmospheric pressure. …
- KEAM 2025Set eng-2025-04294 marksMCQQ.If the acceleration due to gravity on the surface of a planet of mass m and radius r is g, then the escape velocity of a body from the surface of the planet is (A) 2gr (B) r2g (C) gr (D) gr2 (E) 2gr2
›Reveal solutionSolution
Using GM=gr2 in ve=2GM/r gives ve=2gr.
Surface gravity relation. g=r2GM, hence GM=gr2.
Escape velocity. …
- KEAM 2024Set eng-2024-06054 marksMCQQ.If a body is taken above the surface of earth, it looses its weight by 20 % at a height of (A) 25R (B) (25−3)R (C) (25−1)R (D) (25−2)R (E) (1+25)R
›Reveal solutionSolution
A 20% weight loss means g′=0.8g; invert the inverse-square height law to get h.
Above the surface, g′=g(R+hR)2. Losing 20% of weight means g′=0.8g:
(R+hR)2=0.8=54. …
- KEAM 2024Set eng-2024-06064 marksMCQQ.The escape speed of the moon when compared with escape speed of the earth is approximately (A) twice smaller (B) thrice smaller (C) 4 times smaller (D) 5 times smaller (E) 6 times smaller
›Reveal solutionSolution
Escape speed ve=2GM/R; the moon's is about one-fifth of the earth's.
The escape speed is ve=R2GM.
For the earth ve≈11.2 km/s and for the moon ve≈2.38 km/s. The ratio is …
- KEAM 2024Set eng-2024-06084 marksMCQQ.The value of escape velocity ve for a planet depends on (A) the mass of the body thrown from the planet (B) the direction of projection of the body (C) the angle of projection (D) only on the mass of the planet (E) its mass M, density ρ and radius of the planet
›Reveal solutionSolution
ve=2GM/R is independent of the thrown body and its direction; it depends only on the planet's M, ρ and R.
The escape velocity from a planet is
ve=R2GM.
Using M=34πR3ρ, we can also write ve=R38πGρ.
Key points:
- It does not depend on the mass of the body thrown (eliminates A).
- It does not depend on the direction/angle of projection (eliminates B, C). …
- KEAM 2024Set eng-2024-06094 marksMCQQ.If the acceleration due to gravity on the surface of a planet is 2.5 times that on earth and radius, 10 times that of the earth, then the ratio of the escape velocity on the surface of a planet to that on earth is (A) 1:1 (B) 1:2 (C) 2:1 (D) 1:5 (E) 5:1
›Reveal solutionSolution
Escape velocity v=2gR. With gp=2.5ge and Rp=10Re, the ratio is 2.5×10=5, so 5:1.
Escape velocity from a planet's surface:
v=2gR.
Therefore …
- KEAM 2021Set eng-2021-P1-A14 marksMCQQ.If the earth shrinks to half of its present size and its mass reduces to half of its actual mass, then the acceleration due to gravity (g) on its surface will be (A) 4g (B) g (C) 2g (D) 2g (E) 3g
›Reveal solutionSolution
The surface gravity becomes 2g.
Concept and Intuition
Surface gravity is g=GM/R2. Halving the radius quarters the denominator while halving the mass, so the net effect is a factor of 2.
Step-by-Step Solution
- g=R2GM.
- New values: M′=M/2, R′=R/2.
- g′=(R/2)2G(M/2)=R2/4GM/2=24⋅R2GM=2g. …
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