Angular kinematics: determining satellite orbital radius

  • #1
So, I'm working on my semester finals for my high school physics class, and I've run across a problem that has me stumped. I've included the pertinent material below.

Homework Statement


A rocket with a mass of 2kg and traveling at 14,008.34 meters/second loses 10% of its velocity as finishes the remainder of its trip out of earth's gravity field. Calculate the predicted orbital radius for the satellite based on the final velocity of the satellite, given that the mass of the earth is 5.97x1024 kg.

Homework Equations


I feel like the angular kinematics equations would be important here, as would F=ma and some other formulas related to angular kinematics (see below)

Angular kinematics formulas:
ωfi+αt
Δθ=ωit+½αt2
ωf2i2+2αΔθ

where ω is angular velocity, α is change in angular velocity, and t is time.

Other formulas:
F=ma
Vt=rω

where Vt is tangential speed, r is radius and ω is angular velocity.
There might be more formulas that should be here, but I can't think of any off the top of my head.

The Attempt at a Solution


So, my eventual goal is to find the value of r. Since I've been given the mass of Earth, I feel like I need to be doing something involving the gravitational force between the objects. We haven't gone over universal gravitation in our class yet, so I wouldn't imagine its anything to do with that. I'm assuming that the 12,607.506 meters/second I have for a new velocity after accounting for the 10% loss would become the tangential speed of the satellite.

I tried using good old F=ma to calculate the centripetal force, and got a value of 19.6 newtons, given the rocket's mass of 2kg and an acceleration due to gravity of 9.8 meters/second2. Where I'm getting stuck is in translating the knowns that I have (centripetal force of 19.6 newtons, tangential speed of 12,607.506 meters/second) into something I can determine radius with. I always seem to be one unknown short. If I knew the orbital period, it would be trivial to get the radius, knowing that it travels 2π radians in a full rotation.

First of all, are my assumptions correct? Any hints are very welcome. I feel like its some little thing I'm missing that will make all of the pieces come together :)
 

Answers and Replies

  • #2
SteamKing
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The acceleration due to gravity of 9.8 m/s2 is valid only at the surface of the earth. As altitude increases, this value decreases.

Orbits of satellites have everything to do with universal gravitation (indeed as does the determination that g = 9.8 m/s2), so assuming that gravitation is not involved is a curious one.

See this article:

http://en.wikipedia.org/wiki/Orbit

especially the section entitled "Understanding orbits".
 
  • #3
The acceleration due to gravity of 9.8 m/s2 is valid only at the surface of the earth. As altitude increases, this value decreases.

Orbits of satellites have everything to do with universal gravitation (indeed as does the determination that g = 9.8 m/s2), so assuming that gravitation is not involved is a curious one.

See this article:

http://en.wikipedia.org/wiki/Orbit

especially the section entitled "Understanding orbits".
You're right, allow me to clarify. I meant that in the course of our physics lessons, we haven't covered universal gravitation, so I'm assuming this is something that can be solved using our existing knowledge of angular kinematics and rotational motion. Universal gravitation is something that we start covering the beginning of the spring semester, and I can't imagine our teacher requiring us to use something that he hasn't taught yet to solve the problem.

Thanks for the acceleration due to gravity tip, though. It makes sense that it would be different at altitude, so I'm not sure why I assumed it would be the same.
 
  • #4
SteamKing
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I can't imagine our teacher requiring us to use something that he hasn't taught yet to solve the problem.
It wouldn't be the first time this has happened.
 
  • #5
It wouldn't be the first time this has happened.
Well, no, but I've been taking classes (including a full year of biology and chemistry) from this guy for several years now, and he's never given us homework beyond what he's taught in the classes.
 
  • #6
Bandersnatch
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finishes the remainder of its trip out of earth's gravity field. Calculate the predicted orbital radius
I don't see how this makes any sense. It could be a trick question if you had covered gravity, escape velocity and shapes of orbits. But you say you hadn't.
Is the question stated accurately?
 
  • #7
haruspex
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Universal gravitation is something that we start covering the beginning of the spring semester, and I can't imagine our teacher requiring us to use something that he hasn't taught yet to solve the problem.
Does your teacher make a habit of including irrelevant information? You've been given the mass of the Earth. How will you use that except in considering the universal law of gravitation?
 
  • #8
I don't see how this makes any sense. It could be a trick question if you had covered gravity, escape velocity and shapes of orbits. But you say you hadn't.
Is the question stated accurately?
It's part 2 of a two-part question, I've attached the first part in its entirety below (bold text).

Does your teacher make a habit of including irrelevant information? You've been given the mass of the Earth. How will you use that except in considering the universal law of gravitation?
He usually doesn't provide much irrelevant information, but I suppose there is a possibility that it could be in this case.

2) Suppose a person decides to make a low-budget satellite by bundling 1500 D11-5 (model rocket engines that produce 11 newtons of force each and have a 5 second cutoff before falling off) engines together. Calculate the height and final velocity of the rocket at the end of the thrust phase, given the total mass of the rocket is 2kg (assume that the mass doesn't change).

My answers to this question:
Final velocity at end of thrust phase: 14,008.34 meters/second
Final height at end of thrust phase: 11,907.089 meters

3) Assuming that the rocket loses 10% of its velocity as finishes the remainder of its trip out of earth's gravity field, calculate the predicted orbital radius for the satellite based on the final velocity of the satellite, and given that the mass of the earth is 5.97x1024 kg.
 
  • #9
haruspex
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2) Suppose a person decides to make a low-budget satellite by bundling 1500 D11-5 (model rocket engines that produce 11 newtons of force each and have a 5 second cutoff before falling off) engines together. Calculate the height and final velocity of the rocket at the end of the thrust phase, given the total mass of the rocket is 2kg (assume that the mass doesn't change).

My answers to this question:
Final velocity at end of thrust phase: 14,008.34 meters/second
Final height at end of thrust phase: 11,907.089 meters
I don't get those answers. Please post your working. (I get exactly your answers if I plug in 1.7 seconds.)
However, we agree that the height is small compared with Earth's radius, so using 9.8m/s2 throughout is reasonable.
To complete the question, you need the formula for centripetal acceleration. I note you did not include this in Relevant Equations.
 
  • #10
I don't get those answers. Please post your working. (I get exactly your answers if I plug in 1.7 seconds.)
However, we agree that the height is small compared with Earth's radius, so using 9.8m/s2 throughout is reasonable.
To complete the question, you need the formula for centripetal acceleration. I note you did not include this in Relevant Equations.
Ahh, yes, I didn't clarify in my previous posts; there is a chart attached to my homework detailing the thrust of a single D11-5 engine that states that the thrust phase is 1.7 seconds long. The 5 second period is the coasting phase before the parachute deploys, not the period in which thrust is being applied. My bad, sorry for the confusion.

You're right, I don't have centripetal acceleration in the relevant formulas section. I don't believe I tried anything involving that to solve it. Centripetal acceleration is calculated by the formulas ac=Vt/r and ac=rω2, where Vt is tangential speed, r is radius (what we're trying to determine), and ω is angular velocity (if I remember correctly). If my assumption is correct that 12,607.506 meters/second is Vt, then it would seem I can't solve for r without knowing ac. Using the second formula, I would need to know ω, which should be determined by ω=rt, where r is rate and t is time (the angular version of d=rt). Problem is, I don't know time, so it seems that I can't determine centripetal acceleration. I must be missing something here...

Thanks for the help thus far, as well!
 
  • #11
haruspex
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ac=Vt/r
That's not quite right. You left something out.
it would seem I can't solve for r without knowing ac.
But you do know ac. Draw a free body diagram of the craft in orbit. What forces act?
 
  • #12
That's not quite right. You left something out.

But you do know ac. Draw a free body diagram of the craft in orbit. What forces act?
Ahh, you're right, I forgot to square tangential speed. Thanks for the catch :).

I drew a free body diagram and ended up with a centripetal force towards the Earth (the gravitational pull, which, if we are using 9.8m/s2 as our acceleration due to gravity, is equal to 19.6 newtons) and an applied force perpendicular to the centripetal force. Centripetal force is calculated by Fc=m*Vt2/r. Knowing centripetal force, mass, and tangential speed, it seems all I needed to do was solve for r. After doing so, my calculated value of r was 16,219,306.89 meters (I have meters because I was working with only SI units).

To me, this seems a very large value, but given the results of some rough calculations I did to get a basic idea of how this satellite was moving, it doesn't surprise me that it is that large. The velocity that the rocket was traveling at at the end of the second problem that I posted 4 posts back indicates that it is traveling at a minimum of ~28,000 miles per hour, which is significantly more than Earth's escape velocity, which is ~25,000 MPH. Given that speed, I suppose there is a possibility that its not orbiting, but rather on an escape vector.
 
  • #13
SteamKing
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Ahh, you're right, I forgot to square tangential speed. Thanks for the catch :).

I drew a free body diagram and ended up with a centripetal force towards the Earth (the gravitational pull, which, if we are using 9.8m/s2 as our acceleration due to gravity, is equal to 19.6 newtons) and an applied force perpendicular to the centripetal force. Centripetal force is calculated by Fc=m*Vt2/r. Knowing centripetal force, mass, and tangential speed, it seems all I needed to do was solve for r. After doing so, my calculated value of r was 16,219,306.89 meters (I have meters because I was working with only SI units).

To me, this seems a very large value, but given the results of some rough calculations I did to get a basic idea of how this satellite was moving, it doesn't surprise me that it is that large. The velocity that the rocket was traveling at at the end of the second problem that I posted 4 posts back indicates that it is traveling at a minimum of ~28,000 miles per hour, which is significantly more than Earth's escape velocity, which is ~25,000 MPH. Given that speed, I suppose there is a possibility that its not orbiting, but rather on an escape vector.
The mean radius of earth is 6,371,000 meters, so your calculated altitude is quite large. I still don't think you can assume g = 9.81 m/s2 is constant over the range of this rocket.
 
  • #14
Delphi51
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The question asks for the "orbital radius", suggesting that you should assume a circular orbit. This means the rocket does not go straight up but turns horizontal during acceleration so that it gains a large horizontal velocity. Without knowing how much of the thrust is horizontal and how much is vertical, you don't really have enough information to do the problem. Knowing that the orbit is circular provides additional constraints - the height above the Earth surface is related to the orbital radius which is in turn related to the velocity of the orbit. If you know something about the force of attraction between Earth and satellite and some orbital equations you might manage to find an answer. It appears to be beyond the scope of your class, though. If you want to tackle it, I suggest you look up circular orbits in a textbook or on the web, perhaps Wikipedia. The crux of circular orbit physics is that the centripetal force on the satellite is the force of gravity. An energy approach might be more appropriate than the accelerated motion calculation that you did, since there are two dimensions involved in the motion. Work with the energy provided by the rockets and the two kinds of energy the satellite has.
 
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  • #15
haruspex
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my calculated value of r was 16,219,306.89 meters (I have meters because I was working with only SI units).

To me, this seems a very large value, but given the results of some rough calculations I did to get a basic idea of how this satellite was moving, it doesn't surprise me that it is that large. Given that speed, I suppose there is a possibility that its not orbiting, but rather on an escape vector.
Yes, it's a crazy speed. In losing 10% it would have gained 2000km in altitude (less if drag included), well short of a suitable height for orbiting with its remaining speed. If you then drop the constant g assumption then, as you say, it will not enter orbit at all.
The numbers make more sense with 1000 engines instead of 1500.
 
  • #16
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The question asks for the "orbital radius", suggesting that you should assume a circular orbit. This means the rocket does not go straight up but turns horizontal during acceleration so that it gains a large horizontal velocity.
Yes, but the question as posed does not require determining the height it reaches by virtue of its engines, so it's ok to assume that a lot of the speed ends up as horizontal motion (partly perhaps by fins). Clearly the radius is supposed to be determined by the final speed, which can be deduced from the given data. Where it breaks down is that it could not have reached that final height losing only 10% of its speed.
 
  • #17
I've decided to submit the answers I have to my teacher. I'm not fully confident in the validity of the answers, but with the deadline rapidly approaching, I figure its best to just submit the answer and ask for clarification. If needed, I can update the thread with his response later on. Thanks to everyone for your help, it is greatly appreciated!
 

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