# Tag Info

## Hot answers tagged orbital-mechanics

45

Shot answer: Was Sputnik-1 "only for beep" - no, it wasn't :) It was technical test of R-7 as space launcher and test of spacecraft in orbit (athough very simple spacecraft). Also scientists at least tried to make atmosphere research with Sputnik-1. (From my current search results I'm not sure they got much.) Long answer: It's current state of my ...

28

I don't know what the USSR was trying to do with it, but I know what the US Navy did with it. Researchers at the Johns Hopkins University's Applied Physics Lab used the Doppler shift on the 20 MHz tone to determine Sputnik-1's orbit, plus ionospheric electron density and a couple of other things (like a transmitter frequency offset of ~1 kHz from the ...

25

Sputnik 1 was pressurized with nitrogen at 1.3atm. The period of the beeping was tied to a pressure sensor. The logic was being that if anything (such as a micrometeoroid) penetrated the satellite, the change in pressure would detect this and inform the scientists on the ground. This simple test had scientific value for the later programs with living samples ...

23

Is such an orbit even possible? TL;DR: If the Sun wasn't around, yes, such an orbit is possible. But since the Sun is around, such an orbit is impossible. About the name of the orbit Quoting from Emily Lakdawalla, who has a bit more gravitas than some random file blogger, What is a geostationary orbit like at Mars? I have to pause here for a brief ...

10

Your comment on this answer has, I think, led me to understand what you are really asking about. What I am saying is that, presuming the inclination to be ZERO (Plane of orbit parallel to the equatorial plane), the entire plane keeps shifting from pole to pole - parallel to the equatorial plane. You want to move the orbit like this: If that is the right ...

8

Inclination changes are very expensive in terms of delta V/fuel. To do a 90 degree plane change you are thrusting against your current ~7 kilometer a second velocity to bring it to zero and also adding thrust to get a ~7 kilometer per second velocity in the new orbital plane. This is more than it took to get into orbit in the first place, and would require a ...

7

Although a little vexing, the GIF below and the answer it comes from shows that even fixed orbit will eventually pass over all places on a body that rotates underneath it that are at any latitude smaller than the inclination. That being said you have to choose an altitude that spread them out evenly so that you don't have to wait a very long time. That being ...

5

After the fact, yes, Sputnik 1 had some scientific value. But all it did was beep. That is a fact. It had no scientific payloads. No matter how much one sugarcoats the after the fact scientific discoveries inferred from Sputnik-1's orbit, the fact is that the Soviet Union intentionally stripped Sputnik-1 of all of the scientific instruments that the USSR ...

5

There are a number of reasons why this might be the case. There are two main things that a rocket has to do. The first is to get them to the right orbit, and the second is to make sure the payloads get deployed safely. The reasons why such a long wait might include: The satellite needed to be deployed over a specific area to ensure it received sunlight when ...

4

Short answer: probably not, because tidal force resulting from slightly different orbits in spacestation-type distance scales is really small, likely smaller than plenty other disturbances. If you calculate a circular orbit 6700 km from Earth centre and then move extra 100 m radially keeping the original angular speed, resulting force due to orbit "...

3

Both internal pressure and temperature of Sputnik 1 were encoded in the radio signal. Analysis of the radio signals was used to gather information about the electron density of the ionosphere. If the temperature exceeded 50 °C (122 °F) or fell below 0 °C (32 °F), another control thermal switch was activated, changing the duration of the radio signal pulses.[...

3

The choice of an inclination of 51.7° for Salyut 6 (and all other space stations with Russian contribution) is straight forward: It's the lowest inclination that can be reached from Baikonur without crossing the border of China during launch. The low inclination allows for the largest possible payloads due to the additional boost from Earth rotation. All ...

3

To answer the revised question, "does KSP have a software migration path that will allow for n-body physics options" Yes, it has its "mod" framework, which has been used to provide these features for several years.

2

The problem with such tools is that they may be considered a national security risk as the principles are similar to those of missiles. However, there are excellent Master theses and PhD dissertations on the topic of optimal control for landing rockets, such as this MSc thesis and this SpaceX paper. These will explain the equations of motion and part of the ...

2

You don't have to implement the propagator. You can download it from space-track.org, just like you download TLEs. Go to space-track, click "Help", then click "SGP4", and arrive at https://www.space-track.org/documentation#/sgp4 . From there, pick Windows or Linux, and download compiled shared libraries of the latest SGP4 as released ...

2

Yes it is constant along the trajectory as the Jacobi energy is computed using the initial state vector. From Wikipedia's Jacobi integral: In celestial mechanics, Jacobi's integral (also known as the Jacobi integral or Jacobi constant) is the only known conserved quantity for the circular restricted three-body problem. Unlike in the two-body problem, the ...

2

Eigenvalues and Eigenvectors Before specifically addressing the monodromy matrix, it's important to make sure you have a physical understanding of what eigenvalues and eigenvectors represent in general. I highly recommend 3blue1brown's youtube video on this topic: I will distill the important points below. Consider the two-...

2

I'll take a stab a this. A "train detector" needs only to detect if the orbits have the following four characteristics: The two spacecraft are actually in orbit at the same time The two orbits are co-planar The two orbits have nearly the same shape (eccentricity) The two orbits have very nearly the same mean motion (orbits per day) I'll assume ...

2

So, I just gave it a try and have reconstructed the orbit with a numerical propagator. Using a state vector converted from the TLE-Data from the 20th Aug, 18:05:26.676 (yes, I know, not exact, but have to be sufficient*) and a BC I choose so the final re-entry epoch matches (just quick and dirty get a reentry 2h later for a 8 day propagation...). Then I ...

1

You've fully described the first orbit (assuming 2D), and have fixed two of the three parameters of the second orbit, leaving the perigee to float. I assume you only care about hitting the apogee exactly at 60 degrees. The minimum impulsive thrust will occur will occur when the new orbit is tangential to the old orbit, so you are thrusting directly along ...

1

Yes, this is not a full answer, but maybe my thoughts will help you to work out the solution by yourself. Or someone else will connect the loose threads: First of all I recommend another notation: "R" should be reserved for the distance between the object and the orbits focus so I recommend using "Alt" instead. "RA" is common ...

1

Your thoughts are not really naive, they are quite good for own ideas. In deed there is just a small adjustment you have to make: The ground around the train is the sun. What happens when you jump out of a train? First you are "in the air", falling down. That is exactly what happens in interplanetary space. You are falling around sun. What happens ...

1

There are too many variables, some of them infinitely variable, to arrive at the optimal burn plan. And then you need to incorporate that burn plan into something that can be carried out with a reasonably priced computer (US $200K is "reasonably priced) than can handle high level radiation. This is not going to be a state of the art computer. What you ... 1 Intuitively, the velocity impulse at periapsis is $$\Delta v = \int_{t_{p} - \Delta{t}}^{t_p + \Delta{t}} a_\text{drag} \mathrm{d}t$$ where$\Delta{t}$is the duration of the periapsis pass. When the eccentricity is high enough, the periapsis pass happens very quickly and$\Delta{t}$is small; therefore, the$\Delta{v}\$ is small and is not sufficient to ...

1

Edited in response to very constructive criticism from @DavidHammen and @CallMeTom. I agree with them, but I didn't say those things in my initial answer, and I should have. If the only source of data you have is a TLE, then you are starting from a low-quality initial state, which you should expect to be wrong by several kilometers. All a high-quality ...

1

Cerres orbital speed is 17km/sec To make it intercept, you need to do a 5km/sec burn according to http://www.projectrho.com/public_html/rocket/appmissiontable.php Cerres is 9x10^20kg In other words you need to make Cerres lose 1x10^30J Of course, we should reduce that number because you’re using the mass of Ceres itself as reaction mass. Since you propose ...

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