Help! I have no idea how to do this.
The earth's orbit is nearly circular with a radius R=93x10^6 miles (the eccentricity is e=.017). Find the rate at which the earth's radial vector sweeps out area in units of ft^2/s. What is the magnitude of the vector J=rxr` for the earth (in units of squared feet per second)?
One knows the time it takes to go around the Sun.
Note that area swept out is PI*R^2/time
that is area in miles^2/year. Convert that to ft^2/sec
J= r x r'= R*dR/dTheta= R*R dTheta/dtime
= R^2 * 2PI/1year
convert that to ft^2/sec
To find the rate at which the Earth's radial vector sweeps out area, we need to calculate the derivative of the area with respect to time.
The area swept by the radial vector over a small time interval Δt is approximately equal to the area of the triangle formed by connecting the initial position of the radial vector, the final position, and the radius of the Earth's orbit.
The area of a triangle is given by the formula A = 1/2 * base * height. In this case, the base is the length of the radius of the Earth's orbit (R) and the height is the length of the radial vector (r).
Let's denote the swept area as dA and the change in the radial vector's length as dr. Thus, we have:
dA = 1/2 * R * dr
Now, let's differentiate both sides of this equation with respect to time (t):
dA/dt = 1/2 * R * (dr/dt)
We know that the ratio dr/dt is the magnitude of the Earth's orbital velocity, V. Therefore, we have:
dA/dt = 1/2 * R * V
To find the magnitude of the vector J = rxr', we need to take the cross product of the radial vector r and its derivative r'. The magnitude of the cross product is equal to the product of the magnitudes of both vectors multiplied by the sine of the angle between them.
Assuming r is in rectangular coordinates (x, y), and differentiating r with respect to time gives r' = (dx/dt, dy/dt). Hence, J = r x r' = (x, y) x (dx/dt, dy/dt).
Using the properties of the cross product, the magnitude of J can be determined as:
|J| = |r| * |r'| * sin(θ)
where θ is the angle between r and r'. Since both r and r' are vectors in the xy-plane, they are perpendicular to each other, and sin(θ) = 1. Therefore, |J| = |r| * |r'|.
To determine the magnitude of vector J, we need to know the magnitudes of r and r'. The magnitude of r is simply the distance from the origin to the point (x, y), which is given by the equation r = √(x^2 + y^2). The magnitude of r' is equal to the magnitude of the velocity vector V.
Since we do not have information about the specific location of the Earth in its orbit, we cannot determine the exact values of r and r'. However, we can provide you with the general formulas and methodology to find these values when specific information is available.
In summary,
- The rate at which the Earth's radial vector sweeps out area is given by dA/dt = 1/2 * R * V, where R is the radius of the Earth's orbit and V is the magnitude of the Earth's orbital velocity.
- The magnitude of the vector J = rxr' depends on the magnitudes of the radial vector r and its derivative r'. The specific values of these vectors can be determined using the formulas r = √(x^2 + y^2) and r' = V, where (x, y) represents the coordinates of the Earth in its orbit and V is the magnitude of the Earth's orbital velocity.