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Physics: Space and Astrophysics - Kepler's Laws and Planetary Motion

Grade 8IB

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

🔑Concepts

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Kepler's First Law (The Law of Ellipses): All planets move in elliptical orbits, with the Sun at one of the two focal points (foci). An ellipse is characterized by its semi-major axis rr and eccentricity ee.

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Kepler's Second Law (The Law of Equal Areas): A line segment joining a planet and the Sun sweeps out equal areas during equal intervals of time. This implies that the planet travels faster when it is at perihelion (closest to the Sun) and slower at aphelion (farthest from the Sun).

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Kepler's Third Law (The Law of Harmonies): The square of the orbital period TT of a planet is directly proportional to the cube of the semi-major axis rr of its orbit. This can be expressed as T2∝r3T^2 \propto r^3.

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Orbital Velocity: Because of the Second Law, the speed of a planet is not constant. The conservation of angular momentum explains why the planet accelerates as it nears the Sun's gravitational pull.

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Newton's Synthesis: Isaac Newton later proved that Kepler's laws are a direct result of the Law of Universal Gravitation, where the force FF between two masses m1m_1 and m2m_2 is given by F=Gm1m2r2F = G \frac{m_1 m_2}{r^2}.

📐Formulae

T2∝r3T^2 \propto r^3

T12r13=T22r23\frac{T_1^2}{r_1^3} = \frac{T_2^2}{r_2^3}

F=GMmr2F = G \frac{M m}{r^2}

v=GMrv = \sqrt{\frac{GM}{r}}

💡Examples

Problem 1:

A hypothetical planet is located at a distance of 4 AU4 \text{ AU} (Astronomical Units) from the Sun. If Earth's orbital period is 1 year1 \text{ year} at a distance of 1 AU1 \text{ AU}, calculate the orbital period of this hypothetical planet.

Solution:

Using Kepler's Third Law: T12r13=T22r23\frac{T_1^2}{r_1^3} = \frac{T_2^2}{r_2^3} Let Earth be planet 1 (T1=1T_1 = 1, r1=1r_1 = 1) and the new planet be planet 2 (r2=4r_2 = 4). 1213=T2243\frac{1^2}{1^3} = \frac{T_2^2}{4^3} 1=T22641 = \frac{T_2^2}{64} T22=64T_2^2 = 64 T2=64=8 yearsT_2 = \sqrt{64} = 8 \text{ years}

Explanation:

By applying the ratio of the squares of the periods to the cubes of the distances, we find that a planet 4 times further from the Sun takes 8 times longer to complete one orbit.

Problem 2:

A satellite is orbiting a planet. If the distance from the center of the planet is doubled, by what factor does the gravitational force change?

Solution:

Newton's Law of Gravitation states: F=Gm1m2r2F = G \frac{m_1 m_2}{r^2} If the new distance rnew=2rr_{new} = 2r, then the new force FnewF_{new} is: Fnew=Gm1m2(2r)2F_{new} = G \frac{m_1 m_2}{(2r)^2} Fnew=Gm1m24r2F_{new} = G \frac{m_1 m_2}{4r^2} Fnew=14FF_{new} = \frac{1}{4} F

Explanation:

Because the gravitational force follows an inverse-square law, doubling the distance reduces the force to one-fourth of its original value.