You are a scientist exploring a mysterious planet. You have performed measurements and know the...


You are a scientist exploring a mysterious planet. You have performed measurements and know the following things: The planet has radius d, it is orbiting his star in a circular orbit of radius b, it takes time T to complete one orbit around the star, and the free-fall acceleration on the surface of the planet is a. Derive an expression for the mass M{eq}_S {/eq} of the star in terms of b, T, and G the universal gravitational constant.

Simplified planetary motion

The motion planets and celestial bodies was extensively studied in the 16th and 17th centuries. From Kepler's laws, we know that planets follow an elliptic orbit with their star placed in one focus of the ellipse. The gravitational force is directed from the planet to the star.

The two-body problem for the planetary motion is fully explained by solving Newton's equation under the action of the central force,

{eq}\vec{F}_{M_sm}=G\dfrac{M_sm}{r^2}\hat{r} \qquad\qquad (1) {/eq},

where {eq}M_s {/eq} is that mass of the star, {eq}m {/eq} the mass of the planet, {eq}G {/eq} the universal gravitational constant, {eq}r {/eq} the distance between the star and the planet and {eq}\hat r {/eq} the unit vector defining the direction of the force.

A simplified model of the planetary motion considers a circular orbit around the star. In such a case, the orbital velocity becomes constant and the acceleration is centripetal, continuously changing the motion's direction. There is no need to solve the orbit's differential equation and every mechanical quantity can be computed in a few steps.

Answer and Explanation:

Let us use equation (1) for the force between the star (with mass {eq}M {/eq} and the planet with mass {eq}m {/eq}). From Newton's second law in the circular orbit, this force must equal the product of the mass and the centripetal acceleration. In scalar form,

{eq}G\dfrac{M_sm}{r^2}=ma_c=m\dfrac{v^2}{r} \qquad\qquad (2) {/eq}.

For the orbit of the mysterious planet {eq}r=b {/eq}.

The other key information is the orbital period, {eq}T {/eq} which can be written in terms of the orbital length and velocity as,

{eq}T=\dfrac{2\pi b}{v} \qquad \qquad (3) {/eq}.

Solving for the velocity in equation (2) and inserting the result in (3),

{eq}T=\dfrac{2\pi b}{v}=2\pi \sqrt{GM_s b} {/eq}.

From this equation, we can obtain the star's mass,

{eq}\boxed{M_s=\dfrac{T^2}{4\pi^2 Gb}} {/eq}.

Notice that the information about the planet's diameter and its gravitational acceleration on the surface are superfluous. With these data, we can determine the planet's mass or other quantity.

Learn more about this topic:

Kepler's Three Laws of Planetary Motion

from Basics of Astronomy

Chapter 22 / Lesson 12

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