Sorry for reposting this but I'm still confused with the last part...
You have been sent in a small spacecraft to rendezvous with a space station that is in a low circular Earth orbit, say, with a radius of 6,720 km, approximately that of the orbit of the International Space Station. Due to mishandling of units by a technician, you find yourself in the same orbit as the station but exactly halfway around the orbit from it! A smaller elliptical orbit fail unpleasantly. Keeping in mind that the life-support capabilities of your small spacecraft are limited and so time is of the essence, can you perform a similar maneuver that will enable you rendezvous with the station? Find the radius and period of the transfer orbit you should use.
The perihelion is
The aphelion is
The period is
Physics - drwls, Saturday, May 7, 2011 at 12:15am
Apply an impulse to the spaceship in the direction of motion to put it into a larger elliptical orbit. The location that the impulse is applied will remain the perihelion location. Apply enough impulse to make the period 50% longer. After the small spacecraft has made one more orbit, the space station will have made 1.5 orbits and be in the right place.
perihelion: 6720 km
semimajor axis a can be otained from a Kepler relation:
(a/6720)^2 = (1.5)^3
a = 6720*(1.5)^1.5 = 12,345 km
Period = 1.5 x original period (about 135 minutes). You can calculate it from original orbit radius
Thanks. So will the aphelion be the perihelion + 2a = 31,410 km??
In order to calculate the period, won't I need the velocity?
Physics - drwls, Saturday, May 7, 2011 at 1:32am
No. See my answer elsewhere
Physics - Catherine, Saturday, May 7, 2011 at 2:17am
I must be doing something wrong because I typed the answers for a second time and they are wrong, I have only one attempt left, can you please check what I did wrong?
For the period what I did was
P = sqrt(R^3(4pi^2)/GM)
P = sqrt ((6720^3(4pi^2)/(6.674E-11(5.9742E24)))
P = 0.1733 s
0.1733 x 1.5= 0.2600
and for the anphelion I got
aphelion = 2a - perihelion
= 2(12,345)-6720
= 17,970 km
Both answers are wrong...
Physics - bobpursley, Saturday, May 7, 2011 at 5:46am
Your period calculation. How can you think that .17 seconds can possibly be right for orbiting the Earth? Your radius in the numbers is in km, you need it in meters. I get about 311 seconds for period, work it out.
Physics - drwls, Saturday, May 7, 2011 at 7:19am
There is still something wrong with the period calculation. It should be over 90 minutes.
Physics - Catherine, Saturday, May 7, 2011 at 2:59pm
Thanks, I now get 5481 seconds for the period, then I have to multiply that by 1.5 right? I get 8222 s
When you say substitute a for R, you meant for the equation of the aphelion?
Physics - drwls, Saturday, May 7, 2011 at 4:33pm
a is the semimajor axis. Substitute it for R in the equation that relates period to R.
Remember that many orbits with periods longer than 3/2 the circular-orbit period (such 5/4 or 7/6) will also work.. they just take longer to rendezvous.
I'm sorry, I don't understand, you mean this equation: P = 2pi sqrt[R^3/(GM)]
won't "a" be 12,345 km?
So is the period 8222 s?
I still don't get why the ansser for the aphelion is wrong...
8222 s is one possible period that will work. There are others, as i have tried to explain.
For any period P you choose, you can use
P = sqrt(a^3(4pi^2)/GM)
to get the semimajor axis, a.
The perihelion remains the same after the maneuver. The new aphelion distance is
2a - (perihelion)
If these numbers don't work for you, I can't explain why.
Those are my last words on this subject
Just wanted to thank you for answering all my questions even though I was clearly lost during most of this process... I finally got it... Thanks =D
Concept Simulation 3.2 at reviews the concepts that are important in this problem. A
golfer imparts a speed of 30.3 m/s to a ball, and it travels the maximum possible distance
before landing on the green. The tee and the green are at the same elevation. (a) How
much time does the ball spend in the air? (b) What is the longest “hole in one” that the
golfer can make, if the ball does not roll when it hits the green?
To calculate the period of the transfer orbit, you need to use the equation:
P = 2π * sqrt[(R^3)/(GM)]
where P is the period, R is the radius of the orbit, G is the gravitational constant, and M is the mass of the Earth.
In this case, the radius of the orbit is given as 6720 km. However, you need to convert it to meters to use in the equation. Since 1 km equals 1000 meters, the radius is 6720 x 1000 = 6,720,000 meters.
The mass of the Earth is approximately 5.9742 x 10^24 kg, and the gravitational constant is 6.674 x 10^-11 m^3/(kg * s^2).
Now, substitute these values into the equation to find the period:
P = 2π * sqrt[(6720000^3)/(6.674 x 10^-11 * 5.9742 x 10^24)]
P ≈ 311 seconds
So, the period of the transfer orbit is approximately 311 seconds.
Now, to calculate the aphelion, you need to use the equation:
aphelion = 2a - perihelion
where a is the semimajor axis.
The given value of a is 12,345 km. Convert this to meters: 12,345 x 1000 = 12,345,000 meters.
Substitute the values into the equation:
aphelion = 2 * 12,345,000 - 6,720,000
aphelion ≈ 17,970,000 meters
So, the aphelion of the transfer orbit is approximately 17,970,000 meters.
Remember that in this scenario, you are performing a maneuver to rendezvous with the space station, and the transfer orbit should have a period 1.5 times longer than the original orbit. Therefore, you need to multiply the period you calculated (311 seconds) by 1.5:
311 * 1.5 ≈ 467 seconds
So, the period of the transfer orbit that will enable you to rendezvous with the space station is approximately 467 seconds.
Keep in mind that there might be other transfer orbits that take longer or shorter periods to achieve the rendezvous, but this specific calculation gives you a transfer orbit with a 1.5 times longer period.