Prove that for an earth satellite,the ratio of its velocity at apogee to its velocity at perigee is the inverse ratio of its distances from apogee and perigee.
The path of a sattelite around the Earth is elliptical. The perigee is the point where the satellite is nearest to the earth, and the apogee is the point where it is farthest from the earth. Let R(p) and R(a) be the respective distances of these points from the centre of the earth, and V(p) and V(a) be the magnitudes of the velocities at these points. Since vectors of velocities V(a) and V(p) are nornal to the position vectors to these points, the angular momenta are
L(p) = m•V(p) •R(p); L(a)=m•V(a) •R(a). Using the law of conservation of angular momentum L(p) = L(a) => m•V(p) •R(p)= m•V(a) •R(a) =>
V(a)/V(p)= R(p)/R(a).
To prove the given statement for an Earth satellite, let's start by understanding the concepts involved - apogee, perigee, and the velocity of the satellite.
Apogee refers to the point in the orbit of a satellite where it is farthest from the Earth, while perigee is the point where it is closest. The velocity of a satellite refers to its speed in orbit. Now let's proceed with the proof.
1. Let's assume that the satellite's velocity at apogee is Va, and its velocity at perigee is Vp.
2. Similarly, let's assume that the satellite's distance from the Earth at apogee is Da, and its distance at perigee is Dp.
3. To calculate the velocity of a satellite in orbit, we can use Kepler's second law, which states that the area swept out by the radius vector of a satellite in a given time is constant.
4. Since the satellite's distance from Earth changes during its orbit, the time taken to travel from apogee to perigee is the same as the time taken to travel from perigee to apogee.
5. Therefore, the area swept out by the radius vector of the satellite is the same at both apogee and perigee.
6. Given that the area swept out by the radius vector is constant, we can write:
(1/2) Va T = (1/2) Vp T
(Va T) = (Vp T) [Equation 1]
Here, T represents the time taken to complete one orbit.
7. Now, let's consider the distances traveled by the satellite at apogee and perigee during the time T. The distance traveled in a given time is equal to the product of velocity and time.
The distance traveled from apogee to perigee is Da, and from perigee to apogee is Dp. We can write:
Va T = Da [Equation 2]
Vp T = Dp [Equation 3]
8. Dividing Equation 2 by Equation 3, we get:
(Va T) / (Vp T) = Da / Dp
(Va T) / (Vp T) = Dp / Da [Inverse Ratio]
9. From Equation 1, we know that (Va T) = (Vp T), so:
(Vp T) / (Vp T) = Dp / Da
1 = Dp / Da
10. This simplifies to:
Dp = Da
11. Hence, we have proven that the ratio of the satellite's velocity at apogee to its velocity at perigee is the inverse ratio of its distances from apogee and perigee.
Note: In this proof, we have assumed a circular orbit for simplicity. The concept holds true for elliptical orbits as well, but the calculations are more involved.