A body of mass 1 kg is moving under a central force in an elliptic orb...
Calculation of Time Period of Elliptical Orbit
Solution:
Step 1:
First, we need to find the magnitude of the central force acting on the body. We know that the angular momentum of the body is given by:
L = mvr
where m is the mass of the body, v is its velocity, and r is the distance of the body from the center of the orbit.
Step 2:
We can write the velocity of the body in terms of the semi-major and semi-minor axes of the elliptical orbit. We know that the area swept by the line joining the body and the center of the orbit in a unit time is constant.
Step 3:
Using Kepler's second law, we can write:
A/t = (πab)/T
where A is the area swept by the line joining the body and the center of the orbit in a time t, a is the semi-major axis of the elliptical orbit, b is the semi-minor axis of the elliptical orbit, and T is the time period of the orbit.
Step 4:
Substituting the values of a and b, we get:
A/t = π(1000)(100)/T
or
A = (π)(1000)(100)t/T
Step 5:
Now, we can write the velocity of the body as:
v = A/r
where r is the distance of the body from the center of the orbit.
Step 6:
Substituting the values of A and r, we get:
v = (π)(1000)(100)t/(T)(1000)
or
v = (π)(100)t/T
Step 7:
We can now write the magnitude of the central force acting on the body as:
F = mv2/r
or
F = m[(π)(100)t/T]2/1000
Step 8:
We know that the magnitude of the central force is given by:
F = GmM/r2
where G is the universal gravitational constant, M is the mass of the central body, and r is the distance of the body from the center of the orbit.
Step 9:
Equating the above two expressions for F, we get:
m[(π)(100)t/T]2/1000 = GmM/r2
or
[(π)(100)t/T]2/1000 = GM/r3
Step 10:
Substituting the values of r, a, and b, we get:
[(π)(100)t/T]2/1000 = GM/(1000)(100)3
or
T2 = (4π2)(10003)/GM
Step 11:
Substituting the values of G, M, and using the given value of the angular momentum, we get