If an object orbiting the Sun sweeps out of an area of A in a time t1 at the closet point of the object's orbital rotation. How much area does the object sweep out at the farthest point of the object's orbit t2? What is the formula answer?
The object orbiting the Sun sweeps out equal areas in equal time intervals, according to Kepler's Second Law of Planetary Motion. This means that the object will sweep out the same area at both the closest and farthest points of its orbit.
Therefore, the area swept out by the object at the farthest point of its orbit (t2) would also be A.
Mathematically, we can represent this as:
Area at t1 = Area at t2 = A
The area swept out by an object orbiting the Sun is governed by Kepler's Second Law, which states that the rate at which a planet sweeps out equal areas in equal times is constant.
To calculate the area swept out at the farthest point of the object's orbit, we need to know the radius of the orbit at that point. Let's denote the radius at the closest point as r1 and the radius at the farthest point as r2.
Since the time period of the object's orbit is constant, the time taken to sweep out a given area at the closest and farthest points will also be constant. Let's denote the time taken at the closest point as t1 and at the farthest point as t2.
The formula to calculate the area swept out by the object at the farthest point of the orbit is given by:
Area2 = Area1 * (t2 / t1)
Where:
Area1 is the initial area swept out at the closest point (A)
t1 is the time taken to sweep out Area1 at the closest point
t2 is the time taken to sweep out Area2 at the farthest point
So, the formula answer is:
Area2 = A * (t2 / t1)
To derive the formula for the swept area, we can start by considering Kepler's second law, also known as the law of equal areas. According to this law, the rate at which an object sweeps out area in a given amount of time remains constant, regardless of its position in the orbit.
Let's denote the area swept out at the closest point of the object's orbit as A1 and the time taken as t1. Similarly, at the farthest point of the orbit, the area swept out is A2, and the time taken is t2.
Based on Kepler's second law, we can establish the following relationship:
A1 / t1 = A2 / t2
To find the formula for the area swept out at the farthest point of the object's orbit (A2), we can rearrange the equation:
A2 = (A1 / t1) * t2
This formula tells us that the area swept out at the farthest point of the object's orbit (A2) is equal to the ratio of the area swept out at the closest point (A1) and the time taken at the closest point (t1), multiplied by the time taken at the farthest point (t2).
Keep in mind that if the object's orbit is an ellipse and not a perfect circle, this formula will still hold true as long as the time intervals t1 and t2 correspond to the equal areas being swept out.
Please note that this explanation assumes you already know the values of A1 and t1. If you have specific values or further context, you can substitute them into the formula to obtain a numerical answer.