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Gravitation - Gravitational Potential Energy

Grade 11ICSEPhysics

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

πŸ”‘Concepts

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Gravitational Potential Energy (UU) is defined as the work done in bringing a body of mass mm from infinity to a point in the gravitational field of another body of mass MM without acceleration.

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The reference point for zero gravitational potential energy is taken at infinity, i.e., U=0U = 0 at r=∞r = \infty.

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Since the gravitational force is attractive, work is done by the field as the mass approaches the source mass. Consequently, Gravitational Potential Energy is always negative for any finite distance rr.

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Gravitational Potential (VV) at a point is the gravitational potential energy per unit mass: V=UmV = \frac{U}{m}.

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The gravitational force is a conservative force, which means the work done depends only on the initial and final positions, and is related to the potential energy by F=βˆ’dUdrF = -\frac{dU}{dr}.

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When a body of mass mm is raised from the surface of Earth (RR) to a height hh, the change in potential energy is given by Ξ”U=Ufβˆ’Ui\Delta U = U_f - U_i. For hβ‰ͺRh \ll R, this simplifies to Ξ”Uβ‰ˆmgh\Delta U \approx mgh.

πŸ“Formulae

U=βˆ’GMmrU = -\frac{GMm}{r}

V=βˆ’GMrV = -\frac{GM}{r}

Ξ”U=GMm(1Rβˆ’1R+h)\Delta U = GMm \left( \frac{1}{R} - \frac{1}{R+h} \right)

Ξ”U=mgh1+hR\Delta U = \frac{mgh}{1 + \frac{h}{R}}

F=βˆ’dUdrF = -\frac{dU}{dr}

g=GMR2g = \frac{GM}{R^2}

πŸ’‘Examples

Problem 1:

Calculate the work done in lifting a body of mass mm from the surface of the Earth to a height equal to the radius of the Earth RR.

Solution:

The work done is equal to the change in gravitational potential energy Ξ”U=Ufβˆ’Ui\Delta U = U_f - U_i. Initial distance r1=Rr_1 = R and final distance r2=R+R=2Rr_2 = R + R = 2R. Using the formula Ξ”U=GMm(1Rβˆ’12R)\Delta U = GMm \left( \frac{1}{R} - \frac{1}{2R} \right), we get Ξ”U=GMm2R\Delta U = \frac{GMm}{2R}. Substituting GM=gR2GM = gR^2, the work done is W=(gR2)m2R=12mgRW = \frac{(gR^2)m}{2R} = \frac{1}{2}mgR.

Explanation:

To move an object away from Earth, work must be done against the gravitational pull. This work is stored as potential energy. Note that the simple mghmgh formula would give mgRmgR, which is twice the correct value because gg is not constant over such a large distance.

Problem 2:

Find the gravitational potential at a point on the surface of the Earth, given M=6Γ—1024Β kgM = 6 \times 10^{24} \text{ kg}, R=6.4Γ—106Β mR = 6.4 \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:

The gravitational potential VV is given by V=βˆ’GMRV = -\frac{GM}{R}. Substituting the values: V=βˆ’(6.67Γ—10βˆ’11)Γ—(6Γ—1024)6.4Γ—106β‰ˆβˆ’6.25Γ—107Β JΒ kgβˆ’1V = -\frac{(6.67 \times 10^{-11}) \times (6 \times 10^{24})}{6.4 \times 10^6} \approx -6.25 \times 10^7 \text{ J kg}^{-1}.

Explanation:

Gravitational potential represents the potential energy per unit mass. The negative sign indicates that the point is within the gravitational field and work must be done to move a unit mass to infinity.

Gravitational Potential Energy - Revision Notes & Key Formulas | ICSE Class 11 Physics