Minimum Satellite Period problem

Click For Summary

Discussion Overview

The discussion revolves around the minimum satellite period problem, specifically comparing the orbital period of a watermelon dropped through a tunnel through the Earth with that of a satellite in a grazing orbit. Participants explore the implications of different orbital shapes and the effects of Earth's rotation on these periods.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant claims that a watermelon dropped through a tunnel returns in 76 minutes, while a grazing orbit takes 6 minutes longer, questioning the reason for this difference.
  • Another participant challenges the figures presented, suggesting that the period should be the same for both cases, proposing a value of 84 minutes.
  • Some participants discuss the effects of Earth's rotation on the perceived orbital period, suggesting that the minimum orbit time might include the rotation of the Earth.
  • There is a debate about the nature of motion through the tunnel, with one participant asserting it is harmonic oscillation, while another insists it is uniformly accelerated motion.
  • One participant notes that the gravitational force inside the Earth is not constant and varies with distance from the center, which affects the calculations for the period.
  • Another participant mentions that real Earth has variable density, which would impact the acceleration of objects falling through it compared to those in orbit.
  • Some participants speculate about the transit time through different shapes of tunnels, with one suggesting that a hypocycloidal shape would yield a shorter transit time.
  • There are claims that the acceleration due to gravity decreases as one approaches the center of the Earth, contradicting earlier assumptions about constant acceleration.

Areas of Agreement / Disagreement

Participants express multiple competing views regarding the calculations of orbital periods and the nature of gravitational forces within the Earth. The discussion remains unresolved, with no consensus on the correct figures or underlying principles.

Contextual Notes

Participants reference various assumptions, such as uniform density of the Earth and the nature of gravitational force, which may not hold in all scenarios. The discussion highlights the complexity of the problem and the need for careful consideration of these factors.

Eric M. Jones
Messages
11
Reaction score
0
Clever guy that I am, I am writing a small paper on Celestial Mechanics...simplified...really really simplified. However, if I bore a tunnel through the Earth and drop a watermelon through it, the watermelon returns to my hand in 76 minutes. This is an example of an elliptical orbit of e=1

But if I use the same figures and shoot the watermelon around the Earth in a grazing orbit, with eccentricity of zero, the orbital period is 6 minutes longer.

Why?

BTW: I suspect any minimum energy subway tunnel of hypocycloidal shape has 1/2 the transit time of the watermelon, or 38 minutes, or 76 minutes round trip. A hypocycloid where the generating circle diameter is the same as the base circle radius generates a straight line through the center.

Comments?
 
Astronomy news on Phys.org
How did you get these figures? The formula for the period should be the same for both cases. And the value around 84 minutes.
 
Mean Earth Radius=6.37 x10E6 meters
a=9.8 m/sec/sec (-)s=1/2at^2
solve for t: t=19.00 minutes. (time for round trip...76 minutes)
So what happened to the 8 minutes...(or the 6)?

Eric
 
I'm only guessing here, but is it possible that the 84 minutes minimum orbit time includes the Earth's rotation? I.E. Cape Canaveral has moved eastward by 8 minutes of rotation during the orbit time? If you shot up a satellite (to the east to gain velocity), then it would be 76 minutes, +8 minutes (e.g.) before it was overhead.

?
 
Eric M. Jones said:
Mean Earth Radius=6.37 x10E6 meters
a=9.8 m/sec/sec (-)s=1/2at^2
solve for t: t=19.00 minutes. (time for round trip...76 minutes)
So what happened to the 8 minutes...(or the 6)?

Eric

This does not make sense.
The motion through the tunnel is not uniform accelerated. The motion in the tunnel is harmonic oscillation. Write the force acting on the falling body as a function of distance from the center, to see this. Then you can use the equation from that motion.
And for the orbital motion you should use the equations for circular motion.
In the end you'll get the same formula for both periods.
Good luck.
 
>>The motion through the tunnel is not uniform accelerated.

Isaac Newton said it was.

>>Write the force acting on the falling body as a function of distance from the center, to see this.

Okay F=ma. Gravity is considered to be simply a point in the center of the sphere.

>>And for the orbital motion you should use the equations for circular motion.

The circular motion or the elliptical motion have the same formula, and the same orbital time

Eric
 
Eric M. Jones said:
Okay F=ma. Gravity is considered to be simply a point in the center of the sphere.

Careful - that point source rule is only valid above the surface of the earth.

When the falling object is below the surface of the Earth (requires a well or tunnel or some such) at a distance R < RE where RE is the radius of the earth, then all of the Earth that lies between R and RE forms a spherical shell with the object inside it. The gravitational field in the interior of such a shell is zero. So when you're calculating the force on the object, it's as if the mass concentrated at the point in the center of the sphere is only the mass between radius 0 and R.
 
Thanks. The bubble of that thought came to the surface an hour ago.
 
Eric M. Jones said:
>>The motion through the tunnel is not uniform accelerated.

Isaac Newton said it was.
He did not. Maybe you misunderstood him.

Eric M. Jones said:
>>Write the force acting on the falling body as a function of distance from the center, to see this.

Okay F=ma. Gravity is considered to be simply a point in the center of the sphere.
Yes, but F is not a constant for the motion through the tunnel.
Eric M. Jones said:
>>And for the orbital motion you should use the equations for circular motion.

The circular motion or the elliptical motion have the same formula, and the same orbital time

Eric
So why don't you use it then and get the correct value for the period?
I wrote you that the motion through the tunnel and around the circular path have the same period. Unfortunately, none of them is 76 minutes.
I don't know what formula are you using.

Eric M. Jones said:
Clever guy that I am, I am writing a small paper on Celestial Mechanics...simplified...really really simplified.
Albert Einstein is widely quoted as saying:
"Everything should be made as simple as possible, but not simpler." (or variants of the same idea)
You may have simplified a little too much.:biggrin:
 
Last edited:
  • #10
With the approximation of a uniform density of earth, the time would be the same.
However, real Earth has a variable density - the acceleration inside is higher than a linear formula would give, and objects falling through Earth would be a bit quicker than objects in an orbit.

BTW: I suspect any minimum energy subway tunnel of hypocycloidal shape has 1/2 the transit time of the watermelon, or 38 minutes, or 76 minutes round trip. A hypocycloid where the generating circle diameter is the same as the base circle radius generates a straight line through the center.
No - with the approximation of a uniform density of earth, every straight line has the same transit time (about 42 minutes). Other shapes (especially the shapes for the shortest transit time) can have shorter transit times.
 
  • #11
Thanks all for making me smarter.

Eric
 
  • #12
probably not very scientific here but, if you consider that gravity's force is centered on the center of the Earth, then the closer to center the faster you accelerate. you would have to accelerate fast enough to overcome the force that you obtained in that process. But when you start to reach the other side (away from the source) you are gravitating away from the source, which would mean that whatever energy you acquired, you will now be using to produce an "escape velocity". So, the equation from the entrance side, should measure enough energy to balance the exit side. Should equal zero, unless an outside force acts on the object. Unless the force of gravity is different depending on where you are on the Earth. Or altitude, depth, things like that. I'm just a student in this at this time, just a thought.
 
  • #13
october said:
probably not very scientific here but, if you consider that gravity's force is centered on the center of the Earth, then the closer to center the faster you accelerate.
Not if you are inside the Earth. For homogenous Earth the acceleration will decrease as you approach the center.
Even for actual (non-homogenous) Earth, as you go deep enough the acceleration deceases.
 

Similar threads

Graduate Are Retrograde Satellites the Key to Stable and Optimal Orbits?
  • · Replies 5 ·
Replies
5
Views
3K
Undergrad Calculate time (or fraction of orb. period) of sat. in umbra
  • · Replies 2 ·
Replies
2
Views
5K
High School A new quasi-satellite was discovered in April: 2016 HO3
  • · Replies 10 ·
Replies
10
Views
3K
Undergrad Sidereal Periods: Definition, Equations & Examples
  • · Replies 1 ·
Replies
1
Views
4K
Satellite Orbiting: Speed, Period, Altitude Calc.
  • · Replies 2 ·
Replies
2
Views
3K
What Are the Calculations for the Orbit of Vanguard 1 Satellite?
  • · Replies 1 ·
Replies
1
Views
4K
Determine Orbital Period Of Satellite
  • · Replies 10 ·
Replies
10
Views
4K
Question regarding centripetal acceleration/period
  • · Replies 8 ·
Replies
8
Views
2K
Orbital Period of satellite in terms of v and r
  • · Replies 4 ·
Replies
4
Views
3K
Undergrad Position and Velocity in Heliocentric Ecliptic Coordinates
  • · Replies 3 ·
Replies
3
Views
5K