A spacecraft and a satellite are at diametrically opposite...

A spacecraft and a satellite are at diametrically opposite positions in the same circular orbit of altitude $500 \mathrm{km}$ above the earth. As it passes through point $A$, the spacecraft fires its engine for a short interval of time to increase its speed and enter an elliptic orbit. Knowing that the spacecraft returns to $A$ at the same time the satellite reaches $A$ after completing one and a half orbits, determine (a) the increase in speed required, (b) the periodic time for the elliptic orbit.

Key Concepts

Circular Orbit Dynamics

Circular orbit dynamics involve understanding how an object maintains a constant altitude and speed around a primary body. In such an orbit, the gravitational force provides the exact centripetal force needed to keep the satellite in constant motion along a circular path. The orbital speed and period are determined by the balance between gravitational pull and the inertia of the orbiting object, typically expressed by the equation v² = ?/r, where ? is the gravitational parameter and r is the orbital radius.

Elliptical Orbit and Kepler's Laws

Elliptical orbits are characterized by varying orbital speeds and distances from the central body, described by two key parameters: the semi-major axis and the eccentricity. Kepler's laws, particularly the third law, state that the square of the orbital period is proportional to the cube of the semi-major axis (T = 2??(a³/?)). This law is crucial for determining the period of an elliptical orbit after an impulsive maneuver.

Impulsive Burn

An impulsive burn is a short-duration, high-thrust maneuver that instantaneously changes a spacecraft's velocity. This maneuver, when executed at a specific point in orbit, can transition a spacecraft from one trajectory (a circular orbit) to another (an elliptical orbit). The calculation of the required change in speed (?v) involves considering the energy difference between the two orbits, and the vis-viva equation is typically used to assess the new orbital parameters post-burn.

Vis-viva Equation

The vis-viva equation, given by v² = ?(2/r - 1/a), is fundamental in orbital mechanics as it relates the speed of an object at any position in its orbit to the distance from the focal body and the semi-major axis of the orbit. This equation is essential for calculating speeds before and after impulsive burns, and for determining the characteristics of the new orbit when a spacecraft transitions from a circular to an elliptical path.

Orbit Rendezvous and Phasing

Orbit rendezvous and phasing refer to the techniques used to ensure that two or more orbiting objects meet or align at specific points along their paths. This involves calculating the differences in orbital periods and positions so that, after maneuvering, the spacecraft’s orbit is synchronized with that of another object (such as a satellite). In problems like the one given, where timing is critical, these concepts help determine the necessary velocity change and the period adjustments needed for synchronized encounters.

Transcript

00:01 Hello everyone here it is given a spacecraft and satellite are diametrically a position in the same circular orbit of altitude 500 kilometers.

00:19 As it passes through the point a, the spacecraft fires its engine for short interval of time to increase its speed and enter the elliptical orbit, knowing that the spacecraft turns to a at the same time the satellite light reaches to a after completing one and half orbit, we have to calculate increase in speed required, periodic time for the orbitic orbit.

00:51 So we have to calculate first part, increase in speed required, delta v required and b, that periodic time for elliptic orbit t.

01:04 Let us start solving.

01:06 As you know for earth's radius is given 6 .3 7 into 10 to over 6 meter.

01:16 Acceleration due to gravity on the earth surface, 9 .81 meter per second square.

01:23 So value of capital g into m is equal to gr square that is 9 .81 into this value you will get 398 .06 10 to the power 12 meter cube per second square.

01:54 For circular orbit of satellite, radius of the orbit is r plus h and r is the radius of the r6 -3, 70 plus altitude that is 500 km.

02:16 So it will be 68070 or 6 .870 into 10 to the power 6 meter.

02:28 And orbital speed is given by for circular orbit gm, you can write for circular orbit of satellite.

02:53 Calculate the value gm 398 .06 10 to the power 12 and r0 is 6 .870 10 to the power 6.

03:15 So orbital velocity of the satellite you will get 7 .6119 and time period of revelation will be 2 pi r0 upon v0.

03:38 R0, as you know, 6 .87, 10 to the power 6.

03:44 Orbital speed we have measured, 7 .6619, 10 to the power 3.

03:51 So time period of population will be 5 .6708 into 10 to the power 3 seconds.

04:07 For elliptic orbit of sisc is given 3 by 2, tau, that is.

05:15 So time period of elliptical orbit of speed spacecraft is 8 .506 to 10 to the power 3 second.

05:24 So semi -major axis is half of ra plus rv and semi -minor axis is root of ra into rv...

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