Video Transcript
There are two planets in orbit around a star. Both planets have circular orbits. Planet A has a speed of π£, and planet B has a speed of two π£. Give your answers to these questions as a decimal if necessary: what is the ratio of the orbital radii of the two planets, π sub B over π sub A? What is the ratio of the orbital periods of the two planets, π sub B over π sub A?
Letβs start by drawing a sketch of the situation. In this sketch, planets A and B are both in orbit with planet A moving in a circular orbit with instantaneous speed π£ and planet B also in a circular orbit with instantaneous speed two π£. As we look into solving for the ratio of their orbital radii, letβs recall the relationship for the speed of an object in a circular orbit.
In the case of a circular orbit, a satelliteβs speed is equal to the square root of πΊ, the universal gravitational constant, times the mass of the body around which the satellite orbits divided by the radius; that is, the distance from the center of the orbital mass to the satellite.
In our diagram, πB and πA start at the center of the orbited mass. Letβs use this relationship for circular orbit speed to solve for this ratio of radii, the speed of planet A, π£ sub A, equals the square root of capital πΊ times the mass of the star divided by π sub A. And weβre told that that speed is equal to π£. π£ sub B, the speed of planet B, is equal to the square root of capital πΊ times the mass of the star divided by π sub B. Then weβre told that this is equal to two times π£.
If we divide π£ sub A by π£ sub B, then significant cancelation happens in this ratio. We can crossout πΊ and π, the universal gravitational constant and the mass of the star. We can also cancel out the orbital speed, π£. This leaves us with a simplified expression; that one over the square root of π sub A divided by one over the square root of π sub B is equal to one divided by two.
If we then square both sides of the equation, this cancels out the square roots in the left side of our equation and gives us the result one over π sub A divided by one over π sub B equals one divided by four. If we multiply both the numerator and the denominator on the left-hand side of the equation by π sub B, this cancels out π sub B in the denominator, leaving us with a simplified fraction π sub B divided by π sub A which is equal to one-fourth or, in decimal form, 0.25.
That concludes part one. Now we want to move on to solve for the ratio of the orbital periods π sub B over π sub A. As we start here, letβs recall the relationship for orbital period for a circular orbit. Orbital period, π, for a circular orbit equals two π times the square root of the orbital radius cubed all over πΊ times the mass of the planet or object being orbited.
If we write this expression for our scenario, π sub A, the orbital period of planet A, is equal to two π times the square root of π sub A cubed divided by πΊ times π; and likewise, π sub B, the orbital period of planet B, equals two π rimes the square root of π sub B cubed over πΊπ. Just like with the ratio of radii, if we divide π sub A by π sub B, then we see the factors of two π cancel out as well as the factors of one over the square root of πΊ times π.
Weβre left with a ratio π sub A cubed divided by π sub B cubed all raised to the one-half power. Now rather than π sub A divided by π sub B, we want the inverse of that ratio. We can invert both sides of the equation and simplify the exponent so that the right side is π sub B divided by π sub A raised to the to the three-halves power.
Our solution earlier show that this ratio, π sub B to π sub A, is equal to one-fourth. When we raise one-fourth to the three-halves power, we find the result of one-eighth or, in decimal notation, 0.125. This is the ratio of the orbital period of planet B to the orbital period of planet A.
