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Gravitation

Apply concepts from Gravitation to problem-solving. Focus on numerical practice and real-world applications.

2-3 Qs/year45 minPhase 2 · APPLICATION

Concept Core

Newton's Universal Law of Gravitation states that every mass attracts every other mass with a force F = Gm_1 m2m_{2} / r2r^{2}, where G = 6.674 x 101110^{-11} N m2  kg2m^{2} \; kg^{-2} is the universal gravitational constant with dimensional formula [M1  L3  T2M^{-1} \; L^{3} \; T^{-2}], m1m_{1} and m2m_{2} are masses (kg), and r is the distance between their centres (m).
The acceleration due to gravity at Earth's surface is g = GM/R2R^{2} [M0  L1  T2M^{0} \; L^{1} \; T^{-2}] (m/s2s^{2}), where M is Earth's mass and R is Earth's radius. At the surface, g is approximately 9.8 m/s2s^{2}.

The value of g varies with position.
With altitude h: g' = gR2gR^{2}/(R + h)2 (exact), or g' = g(1 - 2h/R) for h << R (inverse-square decrease).
With depth d: g' = g(1 - d/R), which is a linear decrease reaching zero at the centre (d = R).
With latitude λ\lambda: geffg_{eff} = g - R ω\omega2 cos2cos^{2}(λ\lambda), maximum at the poles (λ\lambda = 90 degrees, cos = 0) and minimum at the equator (λ\lambda = 0 degrees, cos = 1). The key distinction is that g decreases linearly with depth but by inverse-square with altitude — the formulas are not the same.

Kepler's three laws govern planetary motion. The First Law states planets orbit in ellipses with the Sun at one focus.
The Second Law (law of areas) states the radius vector sweeps equal areas in equal times, meaning areal velocity dA/dt = L/(2m) = constant.
The Third Law states T2T^{2} is proportional to r3r^{3}: T2T^{2} = (4π24\pi^{2}/GM)r3r^{3}, where T is the orbital period and r is the semi-major axis.

Gravitational potential energy U = -GMm/r [M1  L2  T2M^{1} \; L^{2} \; T^{-2}] (J). The negative sign indicates a bound system; U = 0 at infinity. Gravitational potential V = -GM/r [M0  L2  T2M^{0} \; L^{2} \; T^{-2}] (J/kg) is the PE per unit mass.

Escape velocity vev_{e} = 2gR\sqrt{2gR} = 2GM/R\sqrt{2GM/R} [M0  L1  T1M^{0} \; L^{1} \; T^{-1}] (m/s), approximately 11.2 km/s for Earth. Crucially, escape velocity is independent of the mass of the projected body and the angle of projection — it depends only on the planet's mass M and radius R.
Orbital velocity for a satellite at radius r: v0v_{0} = GM/r\sqrt{GM/r} = gR2/r\sqrt{gR^{2}/r} [M0  L1  T1M^{0} \; L^{1} \; T^{-1}] (m/s), approximately 7.9 km/s near Earth's surface.
The relation between them: vev_{e} = 2\sqrt{2} x v0v_{0}.

For a satellite at orbital radius r: KE = GMm/(2r), PE = -GMm/r, Total energy E = -GMm/(2r). The relation |PE| = 2KE holds, and total energy is negative (bound orbit). If E becomes zero or positive, the satellite escapes.

A geostationary satellite has period T = 24 hours, orbital radius r = 42,164 km from Earth's centre (about 35,786 km above the surface), orbits in the equatorial plane from west to east, and appears stationary relative to the ground.

Solved Numerical 1: At height h = R: g' = gR2gR^{2}/(R + R)2 = gR2gR^{2}/(4R24R^{2}) = g/4 = 104\frac{10}{4} = 2.5 m/s2s^{2}.
At depth d = R/2: g' = g(1 - R/(2R)) = g(1 - 12\frac{1}{2}) = g/2 = 5.0 m/s2s^{2}. So g at depth R/2 (5 m/s2s^{2}) is greater than g at height R (2.5 m/s2s^{2}). The depth formula gives a gentler decrease than the altitude formula.

Solved Numerical 2: Planet with mass M' = 4ME4M_{E}, radius R' = 2RE2R_{E}.
Escape velocity: vev_{e} = 2GM/R\sqrt{2GM/R}.
vev_{e}' = 2G(4ME\sqrt{2G(4M_E}/(2RE2R_{E})) = 2x2GME/RE\sqrt{2 x 2GM_E/{R}_{E}} = 2\sqrt{2} x 2GME/RE\sqrt{2GM_E/{R}_{E}} = 2\sqrt{2} x vev_{e}(Earth) = 1.414 x 11.2 = 15.84 km/s.

Solved Numerical 3: Satellite at h = R (so r = 2R), mass m = 200 kg (R = 6400 km = 6.4 x 10610^{6} m, g = 10 m/s2s^{2}).
Orbital velocity: v0v_{0} = gR2/r\sqrt{gR^{2}/r} = gR2/(2R\sqrt{gR^{2}/(2R}) = gR/2\sqrt{gR/2} = 10x6.4x106/2\sqrt{10 x 6.4 x 10^{6} / 2} = 3.2x107\sqrt{3.2 x 10^{7}} = 5657 m/s = 5.66 km/s.
Time period: T = 2π2\pi r/v0v_{0} = 2π2\pi(2 x 6.4 x 10610^{6})/5657 = 2π2\pi x 12.8 x 10610^{6} / 5657 = 14,225 s = 3.95 hours.
Total energy: E = -GMm/(2r) = -gR2gR^{2} m/(2 x 2R) = -gRm/4 = -(10)(6.4 x 10610^{6})(200)/4 = -3.2 x 10910^{9} J = -3.2 GJ.

The key testable concept is the variation of g with altitude (inverse-square) versus depth (linear), and the relationship vev_{e} = 2\sqrt{2} x v0v_{0}, with escape velocity being independent of the projected body's mass.

Key Testable Concept

The key testable concept is the variation of g with altitude (inverse-square) versus depth (linear), and the relationship v_e = sqrt(2) x v_0, with escape velocity being independent of the projected body's mass.

Comparison Tables

A) Variation of g

FactorFormulaBehaviorAt Extreme Value
Altitude h (exact)g' = gR2gR^{2}/(R+h)2Inverse-square decreaseAt h = R: g' = g/4
Altitude h (approx, h << R)g' = g(1 - 2h/R)Linear approximationAt h = R/2: g' ≈ g(1-1) = 0 (fails)
Depth dg' = g(1 - d/R)Linear decreaseAt d = R (centre): g' = 0
Latitude λ\lambdageffg_{eff} = g - Romega2  cos2  λRomega^{2} \; cos^{2} \; \lambdaDepends on cos2  λcos^{2} \; \lambdaAt pole (λ\lambda=90): gmaxg_{max} = g; At equator (λ\lambda=0): gming_{min}

B) Kepler's Laws

LawStatementMathematical FormNEET Application
First (Orbits)Planets orbit in ellipses with Sun at one focusElliptical orbit geometryIdentifying orbit shape
Second (Areas)Equal areas swept in equal timesdA/dt = L/(2m) = constantVelocity comparison at perihelion vs aphelion
Third (Periods)T2T^{2} proportional to r3r^{3}T2T^{2} = (4π24\pi^{2}/GM)r3r^{3}Comparing periods of different satellites

C) Satellite Parameters

ParameterFormulaDimensional FormulaSI UnitEarth Value (near surface)
Orbital velocity v0v_{0}GM/r\sqrt{GM/r}[M0  L1  T1M^{0} \; L^{1} \; T^{-1}]m/s~7.9 km/s
Escape velocity vev_{e}2GM/R\sqrt{2GM/R} = 2  v0\sqrt{2} \; v_{0}[M0  L1  T1M^{0} \; L^{1} \; T^{-1}]m/s~11.2 km/s
Time period T2π2\pi r/v0v_{0} = 2π  r3/GM2\pi \; \sqrt{r^{3}/GM}[M0  L0  T1M^{0} \; L^{0} \; T^{1}]s~84.6 min (near surface)
Geostationary orbit radiusr = (GMT24π2\frac{GMT^{2}}{4\pi^{2}})^(13\frac{1}{3})[M0  L1  T0M^{0} \; L^{1} \; T^{0}]m~42,164 km from centre

D) Energy of Satellite

TypeFormulaSignRelation
Kinetic energy KEGMm/(2r)PositiveKE = -E
Potential energy PE-GMm/rNegativePE = 2E
Total energy E-GMm/(2r)Negative (bound)E = KE + PE
RelationPE= 2KE

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