New answers tagged orbital-mechanics
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JPL's DESCANSO website links to online books describing spacecraft navigation. Page 4-19 of Volume 2 states "The point-mass Newtonian acceleration plus the point-mass relativistic perturbative acceleration ... is given by Eq. (54) of Moyer (1971)." So JPL was incorporating relativistic effects in its navigation calculations at least as early as then.
I ...
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As far as I know, there has not been a space mission that would have been impossible without a theory of relativistic physics.
It is true that the relativistic effects are clearly visible in GPS clocks. However, if the theory didn't exist, they'd just classify it under "weird observation" and trim the clocks to match ground station clocks. The weird ...
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Supplemental answer confirming @notovny is correct!
While vis-viva gives you the speed, apparently all directions seem to be still possible!
It seems that I've puzzled myself this time.
import numpy as np
import matplotlib.pyplot as plt
from scipy.integrate import odeint as ODEint
def deriv(X, t):
x, v = X.reshape(2, -1)
acc = -x * ((x**2).sum())*...
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I think that the timing of GPS signals was the first real necessity to apply General Relativity to spaceflight, or else precision would be much lower.
According to (Ashby, 1997) and other sources I found the first GPS satellite launched in 1977 was used to prove that General Relativity will have a noticeable effect on the clocks. It turned out that the ...
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There are no limits on the direction.
The Vis-viva equation will give you a speed. Assuming point-masses and sticking to classical mechanics, the Vis-Viva equation does not care at all about what direction you point your velocity in; It's merely an equation based on how the total Orbital Energy (which is the same for all orbits with the same semimajor ...
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Partial answer so far...
For the trajectories in your drawing the objects will miss the Moon. They pass much closer to Earth, so their orbital motion will be substantially faster than the Moon's so they will pass through the interception points days before the Moon does and definitely miss it.
However if those are elliptical orbits with the same semi-...
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Energy is always conserved, but different oberservers will disagree about how much energy there is, and what forms it takes. Also you have to be sure to include the whole system. Let's return to the Phobos example from the linked question, but be a bit more careful.
Suppose what we actually do is use our teraton nukes to split Phobos into two equal halves ...
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Wikipedia covers this.
The 1 in 1410 probability comes from Long-term impact risk for (101955) 1999 RQ36 in 2009 which notes...
The analysis of impact possibilities so far in the future is strongly dependent on the action of the Yarkovsky effect, which raises new challenges in the careful assessment of longer term impact hazards.
With better ...
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I can answer the question of why Semimajor Axis and Eccentricity aren't included in the calculation, at least.
What you have here are the main Keplerian Orbital Elements that define the plane of your orbit around Earth. Every orbit with the same inclination and Longitude of the ascending node lies in exactly the same plane, which intersects the globe of ...
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TL;DR:
Is the barycenter of the solar system dragging me along through the galaxy?
Yep!
What I don't understand is that once I leave the atmosphere of earth and reach the vacuum of deep space, where the earth's gravity is no longer keeping me in orbit
(minor quibble: you're still in an orbit, just not a closed orbit. This will be a hyperbolic ...
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Yes, it is possible. As James K observed in a comment, the surface gravity of Uranus is slightly less than that of Earth, but its mass is 14 times larger. If Earth were orbiting Uranus, it would be a very large moon, but it would still be considered a moon, and thus a moon with a higher surface gravity than its planet.
The reason this is possible is that ...
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Yes, it is.
Given two spherical, uniform, bodies one with mass $m_1$ and radius $r_1$ and the other with mass $m_2$ and radius $r_2$, then the surface acceleration due to gravity will be equal when
$$r_2 = \sqrt{\frac{m_2}{m_1}} r_1$$
For the Moon to have the same surface gravity as the Earth, we can plug in suitable numbers, and you end up with a radius ...
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Given a pair of objects that are gravitationally bound to each other, they will orbit around their common barycenter (center of mass of the system). The object to be most logically deemed the moon will be the one of lesser mass because it will be further from the barycenter than its companion.
For example, Pluto has a gravitationally bound companion named ...
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Gravity isn't just about mass, but about distance, too.
Our moon has a surface gravity of about 1/6th of Earth, because it is small and less dense than the Earth is. Surface gravity of a body is inversely proportional to the square of its radius, holding mass constant. That means that if you compressed the moon such that it was $\frac{1}{\sqrt{6}}$th of its ...
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How can Earth-Centered Inertial (ECI) coordinates be inertial if Earth's orbital motion is always accelerating?
It is true that "Earth-Centered Inertial" is a bit of a misnomer. What this means is that one has to account for the fictitious acceleration that results from the acceleration of the frame of reference. Unlike the fictitious accelerations that ...
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What the heck are "space-fixed coordinates"? To what in space can a coordinate system be fixed?
That's two questions. The answer to the first is that those are Earth-centered inertial or Moon-centered inertial coordinates as indicated by the "Ref. body" column. Look at the velocities. 25600 ft/sec is orbital velocity for a vehicle in low Earth orbit while ...
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Some key points to consider:
The longer you are applying thrust against gravity, the more fuel you require.
An orbit is an orbit; low-Earth orbits, the Moon's path around Earth (or more correctly, the Earth-Moon barycenter), and even Earth-Moon or Moon-Earth transfers are all orbits, and arguably all Earth orbits. Earth-Moon transfer is a special case ...
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Probably interesting: CORONA: America's First Satellite Program
Relevant chunk:
The planned recovery sequence involved a series of maneuvers, each of which had to be executed to near-perfection or recovery would fail. Immediately after injection into orbit, the AGENA vehicle was yawed 180 gegrees so that the recovery vehicle faced to the rear. This ...
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That is how it is commonly used. If you were in a parking orbit around Mars, it is imaginable that the burn that brought you on the transfer back to earth wold be called a TEI. The TEI from Apollo is also a deceleration wrt. to Earth, just like a TEI from Mars would be a deceleration wrt. the Sun.
In 2004, from outside the Earth-Moon system, the Stardust ...
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There isn't a problem with sunlight supply at the south pole. It's actually the opposite situation.
One of the biggest attractions to the lunar south pole (after water) is the fact that there are areas there that are in almost constant sunlight. From the rim of Shackleton crater, the sun skims the horizon to a complete 360˚ as the lunar day progresses. It's ...
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"Yes", sort of, there have been some missions proposed and yes, as you mention, they mainly use solar sails. Actual mission proposals have focused on what reasonable, state-of-the-art solar sails could really do. This limits them to working in places where other orbital forces are easier to counteract.
Probably the one that is furthest along (although, ...
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The answer is no, you cannot have a static satellite. It would require high delta-V continually thrusting the mirror upwards. Solar sails do not have high delta-V.
But that is okay - if all you want is polar moonbase illumination, that problem is already solved:
Solution 1 -
Have a network of satellites with mirrors, and use them in turn as they pass near ...
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Would a lower LEO ISS orbit really have a shorter eclipse duration than a higher one?
No it definitely would not!
Using the equations in the question:
$$f = \frac{1}{2} - \frac{1}{\pi}\arccos \left(\frac{R_E}{R_E+h} \right)$$
$$T = 2 \pi \sqrt{ \frac{(R+h)^3}{GM_E}}$$
where $f$ is the fraction of the orbit in eclipse, T is the period of an Earth orbit ...
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What the other answers fail to mention is that the mass of your orbiting object actually cancels out. It does not matter. See these two equations:
(1) F1 = F2 = G*m1*m2 / r^2
(2) F1 = m1 * a1
Where F is force, G is the universal gravitational constant, m is mass, and r is distance between centers of mass of the orbiting and orbited bodies in question. The ...
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The acceleration due to gravity will be identical regardless of mass, assuming the mass of your spacecraft is negligible compared to mass of the object you're orbiting. For example the Earths moon is large enough to effect the motion of the earth so it doesn't orbit the centre of the earth, but instead it orbits the shared centre of mass of the Earth and ...
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Approximately, yes. The gross gravitational effects on the trajectories of the spacecraft and the other object will be the same.
The force of gravity between two objects is proportional to the product of their masses; by $F = m a$, the acceleration of each object cancels out its own mass ( $a = \frac {F} {m}$ ) and so depends on the mass of the other object....
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This is not a complete answer as I won't be including the exact calculation needed to find out your burn time, but at least I will address the direct return vs bi elliptic approach.
For a return from orbit of a manned spacecraft, you want to balance two factors:
On one side you want to minimise the amount of fuel required for the operation; on the other, ...
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Yes, according to multiple sources, including the answer to this question. The estimated drag forces on the ISS, on average, appear to be about 0.25N (although some estimates put it as high as 0.9N). So yes, in theory, a constant thrust could do it. Now, you'd have to contend with the power drain. I believe HiPEP thrusters use somewhere in the range of ...
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It doesn't matter when, as the system is rotationally symmetric. A retrograde burn is almost certainly the most efficient method (excluding any very long time period wait-for-perturbations-to-become-significant ones).
It is uncommon to de-orbit geostationary object so you might not find much about it directly (the common approach is to increase the orbital ...
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Before getting to the technical feasibility of moving the ISS, I feel obligated to point out that operating it at L1 or lunar orbit is impractical for a few reasons:
The ISS is designed for the radiation environment of low Earth orbit. Outside of low Earth orbit, without the protection of the Van Allen radiation belt, crew aboard the station will receive ...
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The formula is only usable for calculating changes in $a$ in response to small tangential impulses. That is, when the shape of the orbit does not significantly change.
It's an important detail to omit, although $\Delta a$ being proportional to $\Delta v_{t}$ should give a strong hint.
The $20821m/s$ number you get is the cost per meter at a circular LEO ...
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With an elliptical orbit we can't have any fixed Lagrange points, not even the unstable ones aligned with the massive bodies, because the massive bodues are not fixed rekative to each other unless we make a very contrived reference frame. We can, however, conceive of trojan-like objects with the following properties:
*The mean orbit period about a primary ...
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They do of course in the case of the circular orbit, but that is rather trivial. What about elliptical orbits with considerable eccentricity?
Lagrange points aren't really defined for elliptical orbits. They are defined only in the Circular Restricted Three Body Problem (CRTBP or CR3BP). Two bodies have significant masses and the third does not (that's the ...
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According to Wikipedia, the general equation for inclination change is:
$$\Delta{v_i}= {2\sin(\frac{\Delta{i}}{2})\sqrt{1-e^2}\cos(\omega+f)na \over {(1+e\cos(f))}}$$
Where:
$e\,$ is the orbital eccentricity
$\omega\,$ is the argument of periapsis
$f\,$ is the true anomaly
$n\,$ is the mean motion
$a\,$ is the semi-major axis
For circular orbits, this ...
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In his excellent answer Notovy used the vis viva equation to demonstrate same speeds implies r = a.
A basic property points on an ellipse: sum of distance from one focus plus distance to the other focus is 2a.
So if the radius vector has the same length as semi major axis a, that implies the end of the radius vector lies on the end of the ellipse's semi-...
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All right, I'd approach this in the following manner.
Part 1: True Anomaly $\theta$ at $r$ in terms of Elliptical Orbit Eccentricity $e$
The first place I'd start is the Vis-Viva Equation, which, for all orbits around a particular body with a particular gravitational parameter $\mu$ links relative speed $v$ with radial distance $r$ and semimajor-axis $a$.
...
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