Question

In rotational motion, if the radius of orbit is decreased the total energy also gets decreased. Since the universe is an isolated system, the energy should remain constant. Why does such a contradiction is arising?

@ganeshie8 @phi @Irishboy123

Answer 1

if you mean rotational motion intertwined with inverse square attraction, eg grav, then the potential energy falls but you get an increase in kinetic energy or some other trade. so energy or energy/mass is conserved.

dunno if that helps!

Answer 2

\(\color{#0cbb34}{\text{Originally Posted by}}\) @IrishBoy123
if you mean rotational motion intertwined with inverse square attraction, eg grav, then the potential energy falls but you get an increase in kinetic energy or some other trade. so energy or energy/mass is conserved.

dunno if that helps!
\(\color{#0cbb34}{\text{End of Quote}}\)

\(\large \rm E_p=-\frac{GMm}{r} \)
\(\large \rm E_k= \frac{GMm}{2r}\) I can derive this if you want
\[\large \rm E_t=E_p+E_k \]\(\large \rm E_t=-\frac{GMm}{r}+\frac{GMm}{2r} \)
\(\large \rm E_t=-\frac{GMm}{2r} \)
Thus a decrease in radius will result in decrease in total energy.
Why is that?

Answer 3

Yeah, show how you derive the kinetic energy I think that's where the issue is. I'm thinking \(E_k = \frac{1}{2} mv^2\) which I think you are assuming to be uniform circular motion, but in general the orbits can be ellipses which will change that term you've derived. Not 100% sure, just a preliminary guess. I'm working it out right now.

Answer 4

\[\large \rm F_{centripetal}=F_{gravitation}\]
\(\large \rm \frac{mv^2}{r}=\frac{GMm}{r^2}\)
\(\large \rm mv^2=\frac{GMm}{r}\)
\(\large \rm \frac{mv^2}{2}=\frac{GMm}{2r}\)
\[\large \rm E_k= \frac{mv^2}{2}\]\(\large \rm E_k=\frac{GMm}{2r}\)

Answer 5

Yeah, where does \(\frac{mv^2}{r}\) come from? It is derived from circular motion, not elliptical motion.

Here's the quick derivation I know of,

For movement around a circle, position is a constant length (not true for elliptical motion):
\[r\cdot r = C\]
differentiate,
\[v \cdot r = 0\]
velocity is perpendicular to position in circular motion as we know. Differentiate once more,
\[a \cdot r + v \cdot v = 0\]
since we know centripetal acceleration is antiparallel to the position vector plugging in the dot products and replacing the vectors with their magnitudes,
\[-ar+v^2=0\]\[a=\frac{v^2}{r}\]

So this derivation shows you're using uniform circular motion which is not true in general.

Answer 6

The question asks me to consider the orbit of satellite to be in circular motion

Answer 7

Btw I just asked my teacher. He replied the total energy and their expressions are for energies of the satellite not the whole system.

Answer 8

No, that's not the right solution.

Answer 9

If it's circular motion, then the radius doesn't change so there is no contradiction to "changing the radius" since it is a constant for that particular scenario.

Answer 10

When I said the radius is changed, I was referring to resistive forces which can decrease the speed reducing the orbit of the satellite

Answer 11

Changing the radius represents different circular orbits, you can't jump to different distances except in elliptical orbits.

Answer 12

When I said the radius is changed, I was referring to resistive forces which can decrease the speed reducing the orbit of the satellite

Answer 13

What resistive forces? Speed is constant on a circular trajectory with only centripetal acceleration since centripetal acceleration will only change the direction of the velocity while leaving its magnitude constant.

Answer 14

This is the question

"Small resistive forces acting on the satellite cause the radius of its circular orbit to change.

Use your expressions to state, for the satellite, whether each of the following quantities increases, decreases or remains constant
i. Total energy

Answer 15

Oh, this is different since this is no longer a closed system, so there's no contradiction like you describe in your original question haha.

Answer 16

Yes

Answer 17

some of the energy is lost to heat ?

Answer 18

Yes to resistive forces. Earlier I was accounting them into the the total energy of satellite

Answer 19

\[E_{total} = E_{satellite} + E_{space dust}\]
You could then consider the space dust that's resisting it bouncing off at very high speed as it resists the satellite which is where the excess energy had to go in order to keep the system's energy constant.