Planet from Kepler — star mass, velocity and energy

Problem

Planet from Kepler. A planet orbits a star with period $T = 365$ days and average distance $a = 1{,}5\cdot 10^{11}$ m. Calculate (a) the mass of the star; (b) the average orbital velocity; (c) the average kinetic energy per unit mass.

The planet follows an orbit of average radius $a$ around the star.

Solution

Convert the period: $T = 365 \cdot 86400 = 3{,}15\cdot 10^7$ s.

(a) Mass of the star. From Kepler’s third law $T^2 = \dfrac{4\pi^2 a^3}{GM}$ we get $M = \frac{4\pi^2 a^3}{G\,T^2} = \frac{4\pi^2 (1{,}5\cdot 10^{11})^3}{6{,}67\cdot 10^{-11}\,(3{,}15\cdot 10^7)^2}$ $M = \frac{39{,}5 \cdot 3{,}38\cdot 10^{33}}{6{,}67\cdot 10^{-11}\cdot 9{,}95\cdot 10^{14}} \approx 2{,}0\cdot 10^{30}\ \text{kg}$ This is about the mass of the Sun (as expected: $a = 1$ AU, $T = 1$ year).

(b) Average orbital velocity. For a circular orbit $v = \dfrac{2\pi a}{T}$: $v = \frac{2\pi \cdot 1{,}5\cdot 10^{11}}{3{,}15\cdot 10^7} \approx 2{,}99\cdot 10^4\ \text{m/s} \approx 30\ \text{km/s}$

(c) Kinetic energy per unit mass. $\dfrac{E_k}{m} = \dfrac12 v^2$: $\frac{E_k}{m} = \frac12 (2{,}99\cdot 10^4)^2 \approx 4{,}5\cdot 10^8\ \text{J/kg}$

$\ev{M \approx 2{,}0\cdot 10^{30}\ \text{kg};\quad v \approx 30\ \text{km/s};\quad E_k/m \approx 4{,}5\cdot 10^8\ \text{J/kg}}$

Links

Topics: Gravitation Concepts: Kepler’s laws · Uniform circular motion · Kinetic energy Skills: Symbolic setup