suppose the sun were twice as massive a it actually is. What would be the orbital period of a planet at a distance of 10 AU from the sun?
To determine the orbital period of a planet, we can use Kepler's third law of planetary motion, which states that the square of the orbital period is proportional to the cube of the semi-major axis of the orbit.
In this case, we have to consider the scenario where the sun's mass is twice its actual value and calculate the orbital period of a planet located at a distance of 10 AU (astronomical units) from the sun.
Let's break down the steps to find the new orbital period:
Step 1: Determine the current orbital period
Using Kepler's third law, we can relate the current orbital period (P1) and the current semi-major axis (a1) of the planet's orbit to the sun's mass (M) as follows:
P1^2 ∝ a1^3
P1^2 = k * a1^3
We need to find the constant of proportionality (k) for the current scenario.
Step 2: Determine the new orbital period
Now, with the sun being twice as massive (2M), we can relate the new orbital period (P2) and the same semi-major axis (a1) to the new sun's mass (2M) using Kepler's third law again:
P2^2 ∝ a1^3
P2^2 = k * a1^3
Step 3: Find the ratio for mass and apply it to P2 equation
Since the new sun's mass (2M) is twice the actual mass (M), we can use the ratio between the new mass and the actual mass to get the relationship between the new orbital period (P2) and the current orbital period (P1):
P2^2 = (2M / M) * P1^2
P2^2 = 2 * P1^2
Step 4: Calculate the new orbital period
To solve for P2, take the square root of both sides of the equation:
P2 = sqrt(2) * P1
Now, we can substitute the actual value of P1. The orbital period of a planet located at a distance of 10 AU from the sun is approximately equal to 11.86 years or 4,329.6 days.
P2 = sqrt(2) * 4,329.6
By calculating the equation, we find that the new orbital period (P2) would be approximately equal to 6,120.1 days or 16.8 years if the sun were twice as massive.