# Orbital speed problem (1 Viewer)

#### mrpotatoed

##### Active Member

Answer is D, why not A? If Velocity is the square root of (GM/r), if velocity is decreased, wouldn't radius have to increase? I know it would make no sense for it do so, but it seems to go against the formula...

#### Squar3root

##### listening to tapes
D because when its velocity decreases, it slowly decays back to the planet because it does not have a stable orbit

#### mrpotatoed

##### Active Member
yes I know that much... but it goes against the orbital velocity equation which is why I started this thread...

#### Drsoccerball

##### Well-Known Member
View attachment 32196

Answer is D, why not A? If Velocity is the square root of (GM/r), if velocity is decreased, wouldn't radius have to increase? I know it would make no sense for it do so, but it seems to go against the formula...
If it was remaining in orbit then yes but if something is travelling an a specific orbit and it loses speed it doesn't have the speed to rotate in that orbit and falls back to earth slowly

#### Drsoccerball

##### Well-Known Member
If it was remaining in orbit then yes but if something is travelling an a specific orbit and it loses speed it doesn't have the speed to rotate in that orbit and falls back to earth slowly
And besides it needs energy to be put in the higher orbit but you're reducing the energy

#### Librah

He's asking why the equation doesn't match the result. I'm pretty sure he knows what the answer is and why it would be that intuitively.

Think about what orbital speed means. If you had say, at same point the objects velocity at say 10^8 m/s, but the object was still orbiting the center of mass at r, does that contradict the equation?

#### phamtom44

##### Member
Hmm......

The initial mechanical energy or the orbital energy the probe needs to stay in its initial orbit is:

But because its orbital velocity is halved, and thus the orbital energy decreased, the probe no longer has sufficient energy to maintain its current orbit and so it drops down to a lower altitude which it can maintain with the lower energy.

Ironically the satellite moves faster at this lower orbital radius, but the extra kinetic energy is just derived from the gravitational potential energy that it lost, so that it could drop to a lower orbit that it could maintain.

I guess that's what the orbital velocity equation says? v = k/sqrt(R) so, R decreases -> v increase? :S

#### Attachments

• 1.5 KB Views: 53
Last edited:

#### mrpotatoed

##### Active Member
So, the orbital velocity equation only applies when it is in a stable orbit? When the speed is first halved in the question, the orbit is no longer stable, equation no longer applies.. hence it undergoes orbital decay until the KE it gains from losing GPE puts it back into a stable orbit, at which time the equation will apply again?

#### anomalousdecay

yes I know that much... but it goes against the orbital velocity equation which is why I started this thread...
The probe needs to gain potential energy to get into an orbit of larger radius.

So, the orbital velocity equation only applies when it is in a stable orbit? When the speed is first halved in the question, the orbit is no longer stable, equation no longer applies.. hence it undergoes orbital decay until the KE it gains from losing GPE puts it back into a stable orbit, at which time the equation will apply again?
Think of it like this. You just decreased the speed of the probe. Now there are a few things that can arise from this.

Is the probe obtaining potential energy? No.

So what exactly is going on here? We have a decrease in speed. This means a decrease in kinetic energy.

The relationship $\frac{mv^2}{r} = \frac{GMm}{r^2}$ is referring to when the kinetic energy of the probe and potential energy are equal, giving a stable orbit. The relationship you are referring to comes from a derivation using this thought. Hence, you can't apply it to this situation as the rockets have slowed down the probe. However, the kinetic energy has decreased due to the decrease in kinetic energy. If there is a decrease in kinetic energy, then you have to determine the radius r which will allow for a stable orbit. This is where you will find the radius at which there will be a stable orbit.

However, there was no increase in energy applied to the probe. So the probe will not obtain a stable orbit, as the condition for the stable orbit is with a larger radius of orbit.

#### sharoooooo

##### Active Member
Because orbital decay?

#### phamtom44

##### Member
Because orbital decay?
That's precisely what orbital decay is. Due to friction between the probe and the atmospheric molecules, its speed is reduced, and it cyclically undergoes the process described above (but generally its speed doesn't halve so quickly.