The great quality of the Indian mind, its subtlety, has been demonstrated in the recent Chandrayan-3 probe to the Moon.
"What matters is the journey, not the destination." This adage is amply valid for the Chandrayan.
How did they go? With much, much smaller rockets!
Rockets to the Moon, both from US and Russia, are much bigger and far more expensive, and spend a lot more on fuel.
How is it that the same sort of payload has been sent up to the Moon with a lot less energy?
To understand this, we need to go through the new physics, below, where the notion of internal force is applied.
In the case of the Chandrayan, the internal force for the Chandaryan-Earth system is gravity.
Gravity has been used as the internal force to create the extra energy which would have been spent as fuel.
How exactly did this work? First let us see how conventional rockets to the Moon work. They go from low orbit straight to the Moon, accelerating against the Earth's gravity till they reach the escape velocity. As much fuel as required for that, i
consumed. One advantage here is that it is fast work. Disadvantage is large fuel tanks, large fuel requirements, heavier rockets, more pollution, etc.
How did Chandrayan work? They did use up the fuel required to reach low earth orbit, but they had a long way to go after that!
Instead of forcing the rocket up away from the Earth's gravity, they used it to reach higher speeds. So the direction of the thrust was not towards the Earth as it is for conventional rockets. In its orbit, as it was approaching the Earth, the thrust
was applied in the direction of the gravity, or away from the Earth. This increased the velocity of the Chandrayan in its orbit. To repeat, the thrust was with the pull of gravity, not against it.
Now this made the Chandrayan move fast in the opposite direction! Surely this is madness!
Not quite, the escape velocity had not been reached, so the Earth's gravity pulled it back. The orbit now is more elliptical. As it approched the Earth, the rockets could fire again to make it move faster with the Earth's gravitational pull.
This process was repeated, till the orbit became very elliptical, such that at the farthest, the Earth's gravity was weak and the Moon's pull more significant.
Then the rockets fired towards the Moon, and a lot less energy was used to get out of the Earth's orbit.
Brilliant! They can use more payload with the same fuel, with this strategy. Go to Mars!
At the bottom is the physics issue, of internal force, and velocity addition.
It starts with the fact that a body can be moved with internal force.
(Or as any explosion shows, parts of a body.)
That is, within a given geometry there is a system which is independent of other masses or energies. A rocket is a good illustration - the thrust comes from the escaping gas. But the gas escapes the body - in a universe where there is nothing else but
the rocket, its centre of gravity stays exactly where it was. Ditto for any gun firing a bullet. Stop the bullet in the barrel, there will be an explosion, but the gun will go nowhere. The centre of gravity stays put. Such is inertia.
Let us assume we have a "rocket" which gets thrust, but does not eject mass. In other words, it goes by pure internal force. Obviously, that cannot be! The laws of motion will be violated. For every action there has to be an equal and opposite reaction.
Just cannot happen. But for the sake of argument, let us say it does and see to what it leads to.
Some energy E stored within the "rocket" will be expended to move it from inertia and in the absence of any friction or other forces, it will continue to move in a straight line after accelerating over the period over which the internal force acts.
It will have a velocity say v.
Its kinetic energy with respect to the base reference state will be 0.5mv^2. It will be less than the energy E, for there will be losses involved.
Let us now introduce a efficiency k factor. k is greater than 1.
Thus E = 0.5mV^2k
Let the above processes be repeated N times. So with respect to the earliest frame of reference, the velocity of the "rocket" will be Nv. The energy expended will be NE.
The kinetic energy of the "rocket" with respect to the base reference state will be 0.5m(Nv)^2 where m is the mass of the "rocket".
When N is large compared to k, the term
(0.5m(Nv)^2 - NE) will be positive, indicating that extra energy has been created. Let us call it e.
Then
e = 0.5m(Nv)^2 - NE
or e = 0.5mNNv^2 - 0.5mNkv^2
or e = 0.5mNv^2(N-k)
This was derived in my book "To the Stars" published online in my (now defunct) adda website in January 2000.
More details in:
Introduction to "A New Look Towards the Principles of Motion" https://groups.google.com/d/msg/sci.physics/1wmee5C8mFs/kJMPdnFkAwAJ
Section 1
Linear Motion, Momentum, Force, Energy, Internal Force Engines, and the design of Interstellar Spacecraft
https://groups.google.com/d/msg/sci.physics/GbpQC3a2d1Q/jSXQeb9kAwAJ
Section 1 (contd.)
Linear Motion, Momentum, Force, Energy, Internal Force Engines, and the design of Interstellar Spacecraft
https://groups.google.com/d/msg/sci.physics/P9ZiinIDhHU/ZtMQVyliBQAJ
Section 2
The Creation and Destruction of Energy https://groups.google.com/d/msg/sci.physics/wY6_9V8ucSY/3nnJQk9iBQAJ
Cheers,
Arindam Banerjee
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