Question

Two satellites, A and B are in different circular orbits

aboutthe earth. The orbital speed of satellite A is twice that
ofsatellite B. Find theratio (Ta/Tb) of the
periods ofthe satellites.

Answer 1

The ratio of the orbital periods (Ta/Tb) of two satellites, where the orbital speed of satellite A is twice that of satellite B, can be found using Kepler's third law, which relates the orbital period squared to the radius of the orbit cubed. By understanding the relationship between orbital speed and radius, the ratio of the orbital periods can be calculated.

Step-by-step explanation:

When comparing two satellites, A and B, with orbital speeds such that the orbital speed of satellite A is twice that of satellite B, we are tasked with finding the ratio of their orbital periods (Ta/Tb). This problem is grounded in the principles of classical mechanics and specifically relates to Kepler's laws of planetary motion.

According to Kepler's third law, the square of the orbital period (T) is proportional to the cube of the radius of the orbit (r). Mathematically, this is expressed as T² ≈ r³ for a satellite orbiting a much larger body, such as the Earth. Since the gravitational attraction provides the necessary centripetal force for the satellite's circular motion, the gravitational force is also centrally involved in this relationship.

To compare the periods of two satellites, we use this proportionality. If the orbital speed of satellite A is twice that of satellite B, the radius of the orbit is also related to the speed. Specifically, speed is directly related to the square root of the radius of the orbit based on centripetal force considerations. Since the speed of satellite A is twice that of satellite B (VA = 2*VB), it follows that the radius of A's orbit would be four times that of B's orbity (rA = 4*rB). Applying Kepler's law then allows us to find the period ratio as (Ta/Tb)² = (rA/rB)³, and after substituting the relation for the radii, we can solve for (Ta/Tb).

Answer 2

The ratio of the orbital time periods of A and B is
`(1)/(2)`

Solution:

As per the question:

The orbit of the two satellites is circular

Also,

Orbital speed of A is 2 times the orbital speed of B

`v_(oA) = 2v_(oB)` (1)

Now, we know that the orbital speed of a satellite for circular orbits is given by:

`v_(o) = \farc{2\piR}{T}`

where

R = Radius of the orbit

Now,

For satellite A:

`v_(oA) = \farc{2\piR}{T_(a)}`

Using eqn (1):

`2v_(oB) = \farc{2\piR}{T_(a)}` (2)

For satellite B:

`v_(oB) = \farc{2\piR}{T_(b)}` (3)

Now, comparing eqn (2) and eqn (3):

`(T_(a))/(T_(b)) = \farc{1}{2}`