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Fields - Gravitational Fields

Grade 12IBPhysics

Review the key concepts, formulae, and examples before starting your quiz.

πŸ”‘Concepts

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Newton's Law of Universal Gravitation states that the attractive force FF between two point masses MM and mm is proportional to the product of their masses and inversely proportional to the square of the distance rr between them: F = G rac{Mm}{r^2}.

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Gravitational Field Strength (gg) is the force per unit mass exerted on a small test mass placed at a point in the field. For a point mass, g = rac{GM}{r^2} and is a vector pointing towards the mass MM.

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Gravitational Potential (VgV_g) is the work done per unit mass in bringing a small test mass from infinity to a point in the field. It is a scalar quantity defined as V_g = - rac{GM}{r}, where the potential at infinity is zero.

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Gravitational Potential Energy (EpE_p) of a mass mm at a distance rr from mass MM is given by E_p = - rac{GMm}{r}. It represents the work done to move the mass from infinity to that point.

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Equipotential surfaces are surfaces of constant gravitational potential. No work is done when moving a mass along an equipotential surface, and these surfaces are always perpendicular to gravitational field lines.

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Escape velocity (vescv_{esc}) is the minimum vertical velocity required for an object to escape the gravitational pull of a planet and reach infinity: vesc=2GMRv_{esc} = \sqrt{\frac{2GM}{R}}.

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For a satellite in a circular orbit, the centripetal force is provided by gravity: mv2r=GMmr2\frac{mv^2}{r} = \frac{GMm}{r^2}. This leads to Kepler's Third Law: T2∝r3T^2 \propto r^3.

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Total energy of an orbiting satellite is the sum of its kinetic and potential energies: Etotal=Ek+Ep=GMm2rβˆ’GMmr=βˆ’GMm2rE_{total} = E_k + E_p = \frac{GMm}{2r} - \frac{GMm}{r} = -\frac{GMm}{2r}.

πŸ“Formulae

F=Gm1m2r2F = G \frac{m_1 m_2}{r^2}

g=Fm=GMr2g = \frac{F}{m} = \frac{GM}{r^2}

Vg=βˆ’GMrV_g = -\frac{GM}{r}

Ep=mVg=βˆ’GMmrE_p = m V_g = -\frac{GMm}{r}

W=mΞ”VgW = m \Delta V_g

vorbit=GMrv_{orbit} = \sqrt{\frac{GM}{r}}

vesc=2GMRv_{esc} = \sqrt{\frac{2GM}{R}}

T2r3=4Ο€2GM\frac{T^2}{r^3} = \frac{4\pi^2}{GM}

πŸ’‘Examples

Problem 1:

Calculate the gravitational field strength on the surface of Mars. Given: Mass of Mars M=6.42Γ—1023Β kgM = 6.42 \times 10^{23} \text{ kg}, Radius of Mars R=3.39Γ—106Β mR = 3.39 \times 10^6 \text{ m}, and G=6.67Γ—10βˆ’11Β NΒ m2Β kgβˆ’2G = 6.67 \times 10^{-11} \text{ N m}^2 \text{ kg}^{-2}.

Solution:

g=GMR2g = \frac{GM}{R^2} g=6.67Γ—10βˆ’11Γ—6.42Γ—1023(3.39Γ—106)2g = \frac{6.67 \times 10^{-11} \times 6.42 \times 10^{23}}{(3.39 \times 10^6)^2} gβ‰ˆ3.73Β NΒ kgβˆ’1g \approx 3.73 \text{ N kg}^{-1}

Explanation:

The gravitational field strength at the surface of a spherical planet is found by treating the planet as a point mass concentrated at its center and evaluating the field at distance RR.

Problem 2:

A satellite orbits the Earth at a height of 400Β km400 \text{ km} above the surface. If the Earth's radius is 6.37Γ—106Β m6.37 \times 10^6 \text{ m} and mass is 5.97Γ—1024Β kg5.97 \times 10^{24} \text{ kg}, find the orbital speed vv.

Solution:

First, find the orbital radius rr: r=Rearth+h=6.37Γ—106+0.40Γ—106=6.77Γ—106Β mr = R_{earth} + h = 6.37 \times 10^6 + 0.40 \times 10^6 = 6.77 \times 10^6 \text{ m} Now calculate vv: v=GMr=6.67Γ—10βˆ’11Γ—5.97Γ—10246.77Γ—106v = \sqrt{\frac{GM}{r}} = \sqrt{\frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24}}{6.77 \times 10^6}} vβ‰ˆ7670Β mΒ sβˆ’1v \approx 7670 \text{ m s}^{-1}

Explanation:

The orbital speed is determined by equating the gravitational force to the centripetal force requirement for a circular path.

Problem 3:

How much work is required to move a 500Β kg500 \text{ kg} mass from the surface of the Earth to an infinite distance away? (Use M=5.97Γ—1024Β kgM = 5.97 \times 10^{24} \text{ kg} and R=6.37Γ—106Β mR = 6.37 \times 10^6 \text{ m})

Solution:

W=Ξ”Ep=Ep,βˆžβˆ’Ep,surfaceW = \Delta E_p = E_{p,\infty} - E_{p,surface} W=0βˆ’(βˆ’GMmR)=GMmRW = 0 - \left( -\frac{GMm}{R} \right) = \frac{GMm}{R} W=6.67Γ—10βˆ’11Γ—5.97Γ—1024Γ—5006.37Γ—106W = \frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24} \times 500}{6.37 \times 10^6} Wβ‰ˆ3.13Γ—1010Β JW \approx 3.13 \times 10^{10} \text{ J}

Explanation:

Work done is the change in gravitational potential energy. Since the potential at infinity is 0Β J0 \text{ J}, the work required is simply the negative of the potential energy at the surface.

Gravitational Fields - Revision Notes & Key Formulas | IB Grade 12 Physics