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JEE Mains Physics · Formula sheet

Gravitation formulas

15 formulas and 38 common traps for JEE Mains Physics Gravitation, grouped by subtopic.

Full notes with worked examples

Newton's Law, Gravitational Field and Potential

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Newton's law and adding the pulls of several masses

Newton's law of gravitation

F=Gm1m2r2F⃗net=∑iF⃗iF = \frac{Gm_1m_2}{r^{2}} \qquad \vec F_{\text{net}} = \sum_i \vec F_i

Field and potential of a sphere, a shell and a cavity

Uniform sphere of mass M and radius R

r≥R: E=GMr2, V=−GMrr<R: E=GMrR3, V=−GM(3R2−r2)2R3r \ge R:\ E = \frac{GM}{r^{2}},\ V = -\frac{GM}{r} \qquad r < R:\ E = \frac{GMr}{R^{3}},\ V = -\frac{GM(3R^{2} - r^{2})}{2R^{3}}

Potential energy of a system and work to rearrange it

Potential energy

U=−∑pairsGmimjrijΔUR→R+h=mgh1+h/RU = -\sum_{\text{pairs}} \frac{Gm_im_j}{r_{ij}} \qquad \Delta U_{R \to R+h} = \frac{mgh}{1 + h/R}

Common traps

Opposite masses leave only their difference

Masses M and 3M at opposite corners pull a test mass at the centre in opposite directions. The net pull is G(3M − M)m/r² towards 3M. Do the pairs first, then add the two diagonal results at 90°.

A semicircle does not cancel

A full ring gives zero force at its centre. A half ring does not: the components along the diameter cancel, but the ones towards the arc add to 2GMm/(πR²).

Moving mass between two bodies weakens the pull

The total mass stays the same, but the product m₁m₂ falls as the split becomes unequal. An equal split always gives the largest force.

Zero field inside a shell, not zero potential

Inside a uniform shell the field is zero, so the potential does not change. It stays at −GM/R, the surface value. 'Potential is zero inside' is the wrong option.

A cavity ratio is always more than 1

F₁ (whole sphere) is larger than F₂ (sphere with a hole). If your F₁ : F₂ comes out below 1, you have inverted it, and the inverted ratio is usually one of the options.

Measure the cavity's distance to its own centre

The negative sphere sits at the cavity's centre, not at O. Check from the figure whether the cavity is on the particle's side or the far side; the distance changes.

Count the diagonals

A square of masses has six pairs, not four. Leaving out the two diagonal pairs is the most common wrong option.

mgh fails when h is comparable to R

Raising m to height 2R gains (2/3)mgR, not 2mgR. Use mgh/(1 + h/R) unless h is a small fraction of R.

Maximum potential energy usually means maximum size

Gravitational potential energy is negative. When a question asks when it is 'maximum', it means largest in magnitude, that is, most negative.

Acceleration due to Gravity: Surface and Height

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g on the surface: mass, radius and density

Surface gravity

g=GMR2=43πGρRg = \frac{GM}{R^{2}} = \frac{4}{3}\pi G\rho R

g at a height above the surface

g at height h

gh=g(1+h/R)2≈g(1−2hR)  (h≪R)g_h = \frac{g}{(1 + h/R)^{2}} \approx g\left(1 - \frac{2h}{R}\right)\ \ (h \ll R)

Common traps

Same density is not same mass

With density fixed, a bigger planet has MORE gravity (g ∝ R). The 1/R² rule holds only when the mass is fixed. Read which quantity the question keeps constant.

A diameter changes in the same ratio as the radius

'The diameter is reduced to one third' means R becomes R/3, so with the same mass g becomes 9g. Do not halve the ratio because the word was diameter.

From the centre or from the surface?

A body '2R from the surface' is 3R from the centre, so g is g/9. A body '2R from the centre' is at height R, so g is g/4. Read the phrase before you write r.

A height of one diameter is 2R

A point whose height equals the earth's diameter is 3R from the centre, so g there is g/9, not g/4.

The 2h/R rule fails for large heights

At h = R/2 the approximation g(1 − 2h/R) gives zero, which is absurd. Use g/(1 + h/R)² unless h is a few percent of R.

Acceleration due to Gravity: Depth and Rotation

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g at a depth below the surface

g at depth d

gd=g(1−dR)g_d = g\left(1 - \frac{d}{R}\right)

Comparing g at a depth with g at a height

Equal g below and above

1−dR=1(1+h/R)2⇒d≈2h  (h≪R)1 - \frac{d}{R} = \frac{1}{(1 + h/R)^{2}} \quad\Rightarrow\quad d \approx 2h\ \ (h \ll R)

Effect of the earth's spin on g

Effective g at latitude λ

g′=g−ω2Rcos⁡2λg' = g - \omega^{2}R\cos^{2}\lambda

Common traps

Depth is linear, height is inverse square

Going down, g falls in proportion to d/R. Going up, it falls as 1/(1 + h/R)². Using the height formula below the surface, or the depth formula above it, gives a wrong option every time.

g is largest at the surface

g rises from zero at the centre to its peak at the surface and falls on both sides of it. A graph that keeps rising inside and outside, or that is flat inside, is wrong.

d = 2h is only for small distances

The rule comes from the approximation 1 − 2h/R. For a depth of R/2 the matching height is (√2 − 1)R, not R/4. If the distance is a sizeable fraction of R, solve the exact equation.

Both directions reduce g

A statement that g increases as you go down is false: g is largest at the surface. Going up and going down both lower it.

No effect at the poles, most at the equator

Statements that reverse this are a regular trap. At the poles the body sits on the axis and does not go round a circle, so rotation changes nothing there.

The floating day equals a grazing orbit's period

T = 2π√(R/g) is about 84 minutes, the same as a satellite skimming the surface. Both come from setting ω²R equal to g.

Escape Velocity and Energy Conservation

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Escape velocity and how it scales between planets

Escape velocity

ve=2GMR=2gR=R8πGρ3v_e = \sqrt{\frac{2GM}{R}} = \sqrt{2gR} = R\sqrt{\frac{8\pi G\rho}{3}}

Energy conservation for launches and falls

Mechanical energy is conserved

12mv12−GMmr1=12mv22−GMmr2rmax⁡=R1−λ2\tfrac{1}{2}mv_1^{2} - \frac{GMm}{r_1} = \tfrac{1}{2}mv_2^{2} - \frac{GMm}{r_2} \qquad r_{\max} = \frac{R}{1 - \lambda^{2}}

Common traps

Equal escape velocity needs equal M/R

Two planets of different mass can have the same escape velocity only if M/R is the same. Equal MR, or equal M/R², does not do it.

With density given, v_e grows with R, not √R

v_e = R√(8πGρ/3). Doubling the radius at the same density doubles v_e. Writing √(M/R) and forgetting that M grows as R³ gives √2 instead.

Direction does not matter

A body thrown sideways, at 45° or straight up needs the same speed to escape. Escape is an energy condition, and kinetic energy has no direction.

A fall from height R gives √(gR), not √(2gR)

v² = 2gh assumes g stays the same over the whole fall. Over a distance R it does not; energy conservation gives v² = gR.

Escape energy is mgR, not ½mgR

The body must climb from −GMm/R to zero, which is GMm/R = mgR. Half of it, ½mgR, is the kinetic energy of a grazing orbit, a different quantity.

r is from the centre

In GMm/r, a point 3R above the surface has r = 4R. Using the height in place of r is the usual slip.

Satellites: Orbital Speed, Energy and Mutual Orbits

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Orbital speed, period and angular momentum

Circular orbit of radius r

vo=GMrT=2πr3GML=mGMrv_o = \sqrt{\frac{GM}{r}} \qquad T = 2\pi\sqrt{\frac{r^{3}}{GM}} \qquad L = m\sqrt{GMr}

Energy of a satellite and the energy to change orbit

Energies in a circular orbit

KE=GMm2rU=−GMmrE=−GMm2rΔE=GMm2(1r1−1r2)KE = \frac{GMm}{2r} \quad U = -\frac{GMm}{r} \quad E = -\frac{GMm}{2r} \qquad \Delta E = \frac{GMm}{2}\left(\frac{1}{r_1} - \frac{1}{r_2}\right)

Bodies orbiting each other

Binary system

Gm1m2d2=m1ω2r1=m2ω2r2T=2πd3G(m1+m2)\frac{Gm_1m_2}{d^{2}} = m_1\omega^{2}r_1 = m_2\omega^{2}r_2 \qquad T = 2\pi\sqrt{\frac{d^{3}}{G(m_1 + m_2)}}

Common traps

r is R + h

Orbit formulas use the distance from the centre. A satellite 'at a height R' has r = 2R. Putting h in place of r is the commonest error on this page.

Speed does not depend on the satellite's mass; L does

Two satellites in the same orbit have equal speed and period whatever their masses. Their angular momentum and energy scale with mass.

PE is 2E, not E/2

Since E = −GMm/2r and U = −GMm/r, U = 2E. A statement that says PE is half the total energy is false.

Higher orbit: more energy, less speed

Raising a satellite needs energy, yet its kinetic energy falls. The extra energy, and more, goes into potential energy.

Launching is not the same as changing orbits

From the ground at rest, the start energy is −GMm/R with no kinetic energy, so ΔE = GMm(1/R − 1/2r). The orbit-to-orbit formula uses −GMm/2r at both ends.

The separation is 2r, not r

Two equal masses on a circle of radius r are 2r apart. Using r in Gm²/r² makes the speed twice too large.

Two different distances in one equation

In a binary, the force uses the separation d; the centripetal term uses the star's own radius about the centre of mass. Each star's radius is the OTHER star's share of the total mass times d.

Kepler's Laws of Planetary Motion

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Third law: period and orbit radius

Kepler's third law

T2=4π2GMr3T2T1=(r2r1)3/2M1M2T^{2} = \frac{4\pi^{2}}{GM}r^{3} \qquad \frac{T_2}{T_1} = \left(\frac{r_2}{r_1}\right)^{3/2}\sqrt{\frac{M_1}{M_2}}

First and second laws: ellipses and equal areas

Kepler's second law

dAdt=L2m=constantvmax⁡rmin⁡=vmin⁡rmax⁡\frac{dA}{dt} = \frac{L}{2m} = \text{constant} \qquad v_{\max}r_{\min} = v_{\min}r_{\max}

Common traps

The central mass matters; the planet's mass does not

T depends on the mass being orbited. A heavier planet in the same orbit has the same period. When two different stars or planets are involved, include √(1/M).

Height or radius?

Kepler's law uses the distance from the centre. A geostationary satellite 6R above the surface is at r = 7R.

Raise the ratio to the power 3/2, not 3

Halving the radius divides T by 2√2, not by 8. Square both sides only after you have written T² ∝ r³.

Fastest when nearest

A statement that a planet is slowest near the sun is false. Constant angular momentum means a smaller r needs a larger speed.

Constant areal velocity, not constant speed

The area swept per second is fixed; the distance travelled per second is not. 'The linear speed is constant' is the incorrect statement in this pair.

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