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A body of mass 1 kg is moving under a central force in an elliptic orbit with semi major axis 1000 m and semi minor axis 100 m. The orbital angular momentum of the body is 100 kg m2 s -1 . The time period of motion of the body is ?
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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
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A body of mass 1 kg is moving under a central force in an elliptic orbit with semi major axis 1000 m and semi minor axis 100 m. The orbital angular momentum of the body is 100 kg m2 s -1 . The time period of motion of the body is ?
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A body of mass 1 kg is moving under a central force in an elliptic orbit with semi major axis 1000 m and semi minor axis 100 m. The orbital angular momentum of the body is 100 kg m2 s -1 . The time period of motion of the body is ? for IIT JAM 2024 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about A body of mass 1 kg is moving under a central force in an elliptic orbit with semi major axis 1000 m and semi minor axis 100 m. The orbital angular momentum of the body is 100 kg m2 s -1 . The time period of motion of the body is ? covers all topics & solutions for IIT JAM 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A body of mass 1 kg is moving under a central force in an elliptic orbit with semi major axis 1000 m and semi minor axis 100 m. The orbital angular momentum of the body is 100 kg m2 s -1 . The time period of motion of the body is ?.
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