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Inter Planetary Flights

Let us consider space flight from one planet to the other, say from earth to Mars. To a good approximation, we shall consider the orbits of the two planets to be concentric circles with Sun at the center.

It turns out that in terms of energy, the most effective method to transfer the rocket from earth to Mars is not to send it directly along the radial lien but to let it go in an elliptic orbit (with Sun at one focus) which joins smoothly to the orbits of earth and Mars. The path of the rocket starts tangentially earth’s orbit at E and joins the orbit or Mars tangentially at M. Note that the minimum distance rmin of the elliptic orbit of the rocket is equal to radius R_M of mars’ orbit, about Sun.

Hence, the eccentricity and orbital parameter of the rocket’s orbit are given as

Taking R_M = 1.5 RE = 225 × 10⁶ km, R_E = 150 × 10⁶ km, we have e = 0.2 and p = 180 × 10⁶ km.

Now, the semi-major axis a of the elliptic orbit is

The energy of the rocket in above trajectory (which remains conserved) is given by

where m is the mass of the rocket.

Now, when the rocket starts its journey in space, it is at the perihelion of its orbit so that its energy is given by,

where v₁ is the initial velocity needed by the rocket; potential energy refers to potential energy due to gravitational force of Sun, at distance R_E from Sun. That means we are assuming that the rocket has already escaped earth’s gravitation.

From conservation of energy, E₁ = E, and hence we find

When rocket finally reaches near the planet Mars, it is at the apehelion of its orbit. Its velocity v₂ there, is given by,

Hence, v₂ = 21.9 km/s.

However, the orbital speed of mars around Sun is,

Thus, in order to tangentially approach the planet Mars, the rocket requires a final impulse which should increase its speed from v₂ to v_M i.e. a small increase by about 2.6 km/s. Once it catches up with the planet, it would land on it guided by local gravity of the planet. Similarly, to begin with, it needs a speed slightly more than the orbital speed of earth, v₁ – v_E = 2.8 km/s. Thus, once taken from earth’s gravity, if it is launched in the direction of earth’s orbital motion, the rocket needs a small initial impulse to get a speed (relative to earth) of about 2.8 km/s. All this requires only a small amount of energy in rocket propulsion.

How much time the rocket takes to reach the planet? This can be found using Kepler’s third law, T² ∝ a³. Hence,

= 1.4 year (approx.)

The journey takes half time period of rocket’s orbit, i.e. it would take nearly 256 days to reach Mars along tangential elliptic orbit. Such a transfer of rocket along an elliptical orbit is called Hohmann transfer in space research.