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Gravitation - Kepler's Laws of Planetary Motion

Grade 11ICSEPhysics

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

πŸ”‘Concepts

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Kepler's First Law (Law of Orbits): All planets move in elliptical orbits with the Sun situated at one of the two foci of the ellipse. An ellipse is defined by its semi-major axis aa and eccentricity ee.

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Kepler's Second Law (Law of Areas): A line segment joining a planet and the Sun sweeps out equal areas during equal intervals of time. This means the areal velocity dAdt\frac{dA}{dt} is constant, which is a direct consequence of the conservation of angular momentum LL.

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Kepler's Third Law (Law of Periods): The square of the time period of revolution TT of a planet is directly proportional to the cube of the semi-major axis rr of its orbit: T2∝r3T^2 \propto r^3.

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Areal Velocity and Angular Momentum: The constant areal velocity is expressed as dAdt=L2m\frac{dA}{dt} = \frac{L}{2m}, where LL is the angular momentum and mm is the mass of the planet.

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Perihelion and Aphelion: At the perihelion (closest point to the Sun, distance rpr_p), the planet moves at its maximum velocity vpv_p. At the aphelion (farthest point, distance rar_a), it moves at its minimum velocity vav_a. By conservation of angular momentum, mvprp=mvaram v_p r_p = m v_a r_a.

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Newton's Deduction: Newton used Kepler's Third Law for circular orbits to deduce the Inverse Square Law, showing that the gravitational force FF is proportional to 1r2\frac{1}{r^2}.

πŸ“Formulae

T2∝r3T^2 \propto r^3

T12T22=r13r23\frac{T_1^2}{T_2^2} = \frac{r_1^3}{r_2^3}

dAdt=L2m=constant\frac{dA}{dt} = \frac{L}{2m} = \text{constant}

vprp=varav_p r_p = v_a r_a

T=2Ο€r3GMT = 2\pi \sqrt{\frac{r^3}{GM}}

πŸ’‘Examples

Problem 1:

The distance of a planet from the Sun is 44 times the distance of the Earth from the Sun. Calculate the time period of the planet's revolution around the Sun in years.

Solution:

Given rp=4rer_p = 4r_e and Te=1T_e = 1 year. According to Kepler's Third Law: Tp2Te2=rp3re3\frac{T_p^2}{T_e^2} = \frac{r_p^3}{r_e^3} Tp212=(4rere)3\frac{T_p^2}{1^2} = \left(\frac{4r_e}{r_e}\right)^3 Tp2=43=64T_p^2 = 4^3 = 64 Tp=64=8Β yearsT_p = \sqrt{64} = 8 \text{ years}

Explanation:

By applying the ratio of the squares of the periods to the cubes of the orbital radii, we can determine the unknown period relative to Earth's orbital parameters.

Problem 2:

A planet is at a distance of r1r_1 from the Sun and moves with a speed v1v_1. When it reaches a point at a distance r2r_2, what will be its speed v2v_2?

Solution:

Based on Kepler's Second Law and the conservation of angular momentum: L1=L2L_1 = L_2 mv1r1sin⁑(90∘)=mv2r2sin⁑(90∘)m v_1 r_1 \sin(90^\circ) = m v_2 r_2 \sin(90^\circ) (Assuming values at perihelion/aphelion or average motion) v1r1=v2r2v_1 r_1 = v_2 r_2 v2=v1r1r2v_2 = \frac{v_1 r_1}{r_2}

Explanation:

Since the gravitational force is a central force, angular momentum is conserved. Therefore, the product of the radial distance and the transverse velocity remains constant throughout the orbit.

Kepler's Laws of Planetary Motion - Revision Notes & Key Formulas | ICSE Class 11 Physics