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Gravitation - Kepler's Laws

Grade 11CBSEPhysics

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

🔑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 foci of the ellipse.

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Kepler's Second Law (Law of Areas): The line that joins any planet to the Sun sweeps out equal areas in equal intervals of time. This means the areal velocity dAdt\frac{dA}{dt} is constant.

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The Law of Areas is a direct consequence of the Law of Conservation of Angular Momentum (LL), as the gravitational force is a central force.

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Kepler's Third Law (Law of Periods): The square of the time period of revolution (TT) of a planet is proportional to the cube of the semi-major axis (RR) of its orbit.

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For circular orbits, the radius RR is used, and the law is expressed as T2∝R3T^2 \propto R^3.

📐Formulae

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

T2∝R3T^2 \propto R^3

T2=(4π2GMs)R3T^2 = \left( \frac{4\pi^2}{GM_s} \right) R^3

L=mvrsin⁡θ=constantL = mvr \sin \theta = \text{constant}

vprp=varav_p r_p = v_a r_a (where vp,rpv_p, r_p are velocity and distance at perihelion and va,rav_a, r_a at aphelion)

💡Examples

Problem 1:

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

Solution:

Given Rp=4ReR_p = 4R_e. According to Kepler's Third Law, T2∝R3T^2 \propto R^3. Therefore, (TpTe)2=(RpRe)3\left(\frac{T_p}{T_e}\right)^2 = \left(\frac{R_p}{R_e}\right)^3. Substituting the values: (Tp1)2=(4ReRe)3=43=64\left(\frac{T_p}{1}\right)^2 = \left(\frac{4R_e}{R_e}\right)^3 = 4^3 = 64. Taking the square root, Tp=64=8T_p = \sqrt{64} = 8 years.

Explanation:

The time period of a planet increases with the orbital radius following the T2∝R3T^2 \propto R^3 relationship.

Problem 2:

A planet moves in an elliptical orbit around the Sun. If its maximum and minimum distances from the Sun are rmaxr_{max} and rminr_{min} respectively, find the ratio of its maximum speed to its minimum speed.

Solution:

By the conservation of angular momentum, L=mvmaxrmin=mvminrmaxL = m v_{max} r_{min} = m v_{min} r_{max}. Therefore, vmaxvmin=rmaxrmin\frac{v_{max}}{v_{min}} = \frac{r_{max}}{r_{min}}.

Explanation:

Since angular momentum is conserved, the planet moves fastest when it is closest to the Sun (perihelion) and slowest when it is farthest (aphelion).

Kepler's Laws - Revision Notes & Key Formulas | CBSE Class 11 Physics