38

It will not. Once released, the object is in a very slightly different orbit from the space station. If you put the object further out or further in from Earth than the space station it will be slightly further from, or slightly closer to Earth, while moving at the same velocity as the Space station, which will result in a slightly more, or less elliptical ...


35

You might like to take a look at Electromagnetic Tethers. A spacecraft moving through a magnetic field (such as the Earth's) can deploy a long, trailing wire and run a current through it. This will provide an electromotive force to the wire (and craft) that can be used to raise or lower orbit. It isn't exactly the same as your idea, but it is a practical ...


34

After writing my comments, I started writing a new answer. That got long, so here's a shorter one. The "energy of an orbit" may be poorly defined and depending on the definition, is not subject to energy conservation laws. Let's ignore it and focus on the kinetic energy. Kinetic energy is not necessarily conserved. Momentum is. Momentum (derived ...


28

Yes, but only because the Earth rotates. If you throw the ball, you will end up in a slightly different spot. If you were at a high point at the equator and threw the ball due East at higher than orbital velocity, you would rotate by the time the ball came back and it would be slightly over your head, because you have moved to a different spot in the orbit. ...


26

Didymos will be roughly 11 million kilometers (6.8 million miles) from Earth at the time of the DART impact.The round trip time for a radio signal from DART to the control center on Earth will be about 73 seconds. The impact speed to Didymos will be about 6.6 km/s, after the 73 seconds DART will be 484 km closer to its target. So DART should navigate ...


25

A first-order approximation to see whether an orbit could be stable is to calculate the radius of the Hill sphere of the parent: $$R_H \approx a\sqrt[3]{\frac{m}{3M}}$$ With the mass or the ISS ($m$), the mass of the Earth ($M$), and the orbital radius of the ISS ($a$), we see that the region the ISS dominates gravitationally has a radius of less than 2 ...


24

The lowest orbit achieved would probably be PFS-2, a small satellite deployed from Apollo 16's service module. It was intended to go into a 55x76-mile orbit (88.5x122 km), but due to variations in the Moon's gravity field, it made passes of six miles (9.6 km) or less before crashing into the Moon's surface. There are very few stable low orbits around the ...


23

What's the root cause of the disparity between what the article says and what's shown in the video? There is no disparity. The article says you need a minimum velocity of two inches per second [bold emphasis mine]: The engineer added, “Simple trigonometry led to the conclusion that pushing an object away at two inches per second within a 30-degree cone ...


23

Lagrange points... can be orbited by asteroids, satellites, and any other useful or interesting object. Assuming two-body motion however... First a quick note; the stability of a Lagrange point itself is not a predictor of the stability of an orbit associated with ("around") it. Are some Halo Orbits actually Stable? (stable orbits about unstable ...


19

"Lowest possible lunar orbit..." As pointed out in comments and in answers to the linked questions Are low, polar lunar orbits in general relatively stable? Moon orbit station-keeping delta-V budget What's the floor for stable retrograde lunar orbits? Besides Luna, what celestial bodies exhibit lumpy gravity? very close orbits around any body ...


19

When we consider the mass something is orbiting about, we assume a gravitational potential corresponding to an inverse-square force, which corresponds to either a point mass or, by Newton's shell theorem, a spherically symmetric mass. It is this situation where any 2-body orbital dynamics can be reduced to the Kepler laws, with periods depending on the ...


18

The sum of all mechanical energy will be the same after your ideal burns. The difference will be that the portion given to your exhaust will be greater for higher burns. For a burn farther away from the planet: The craft will be higher, so the PE of the exhuast will be greater. The craft will (usually) be slower, so the KE of the exhaust will be greater.


16

There is no orbit around the Earth that remains in permanent shadow. Such an orbit would need to have a period of one year, and that orbit would be too large to fit within the Earth's Hill Sphere, which represents a broad upper bound on the range of stable orbits around Earth. In theory, an object near Earth-Sun L2 (at roughly 1.5 million km from Earth) ...


16

Exactly the same way you avoid collisions when not altering orbit. Altering your orbit does not significantly alter your risk of collisions, other than possibly moving you to a higher or lesser densely populated part of orbital space. Orbits are not neat stacks of perfect circles around the planet. All orbits are ellipses, with the perigee closer to the ...


16

The force on the electromagnet will be given by its magnetic moment times the Earth's magnetic field gradient. A large electromagnet might have a magnetic moment of the order of $10^{+6} \,\mathrm{Amp}\cdot\mathrm{meter}^2$. The Earth's field is of the order of $10^{-4} \,\mathrm{Tesla}$ and, at an orbital radius of say $10^{+7}\,\mathrm{meters}$, the field ...


15

Yes, it takes roughly the same amount of delta-v, your analysis is sound and good. But no, there is no error in the book because that's the premise of that part of the story: we can either reach Mars with food and supplies to last Watney until Ares 4 comes around and picks him up OR we resupply Hermes and they pick him up now(ish). The difference is not in ...


14

It's theoretically possible; the velocity of the exhaust plume is around 3000 m/s (pretty close to what you'd need for a translunar injection!) and the mass flow rate is ~270 kg/s, so if a small piece of debris fell off the stage into the plume, it could get quite a boost. It seems a little unlikely that a piece big enough to track would get kicked up in ...


14

Orbital mechanics always apply. For shuttle the two different operational phases were referred to as Rendezvous Ops and Prox Ops The breakpoint between the two was defined in the Space Shuttle Flight Rules, Rule A2-116 (emphasis mine) A. RNDZ OPS ARE DEFINED TO INCLUDE ALL ORBITER RNDZ MANEUVERS AND ASSOCIATED RNDZ ACTIVITIES TERMINATING WITH THE INITIATION ...


13

note: This is a historical answer, and explains how to find tiny asteroids close by. The OP has clarified they want a profitably mineable-sized asteroid so there is room for more answers. Previous wording: ...specific asteroids and/or their asteroid orbit type... As discussed in Have there been any documented mini-moons since 2006 RH120?, "mini-moons&...


13

I think the maximum velocity change from a flyby would help quantify this. $$\Delta v \leq \sqrt{\frac{GM}{r_P}}$$ That is, with perfect relative velocity and angle, the change in velocity from such a flyby perturbation is limited by the mass ($M$) of the asteroid, and the distance at closest encounter ($r_P$), which in the best case is to barely miss ...


13

From Why was the 100m Green Bank dish needed together with DSN's 70m Goldstone dish to detect Chandrayaan-1 in lunar orbit? above: "This computer-generated image depicts the Chandrayaan-1's location at time it was detected by the Goldstone Solar System radar on July 2, 2016. The 120-mile (200-kilometer) wide purple circle represents the width of ...


13

Electromagnets pulling against the Earth's magnetic field are used for the (much lower energy) task of orienting satellites, pointing them in the right direction. This is called a magnetorquer. As another answer suggests, it's not going to be easy to get any substantial delta-V that way.


12

See pp. 14 & 15 in this FAA guide for an explanation, including these great diagrams: I also recommend sketching it out on a roll of paper:


12

I've got a set of Keplerian orbital elements $e_0$, $a_0$, $i_0$, $\omega_0$, $\Omega_0$, and $\theta_0$, and I'd like to get to a different orbit with orbital elements $e$, $a$, $i$, $\omega$, $\Omega$, and $\theta$. How do I calculate (a) the amount of delta-v I'll need for this maneuver or set of maneuvers, and (b) which maneuver or set of maneuvers I ...


11

You're correct. On a perfectly spherical, atmosphere-free Earth, with no obstacles as tall as you, with a uniformly spherical gravitational field, it would be possible; the low point of the orbit would be at the altitude you threw the ball from, a couple of meters above the surface.


11

No, such a calculation is not possible. The performance to different orbits is determined by the specific parameters of each launcher, e.g. specific impulse of each stage engines, fuel ratio, time of staging, flight profile... Have a look at this plot, taken from https://elvperf.ksc.nasa.gov As you can see the two launchers have the same performance to a ...


11

Using Perifocal polar coordinates, where the x-axis points from the central body to the periapsis, and the polar equation for conic sections: $$r=\frac{a(1-e^2)}{1+e \cos(f)}$$ Provided parameters $\mu$ Standard Gravitational Parameter of the central body $r_p$ Periapsis Distance (Point P in the diagram) $v_p$ Speed at Periapsis $r$ Radial distance at ...


10

If we assume Keplerian/Newtonian mechanics, then we can see a way to rendering the same local curvature of the path at perigee and at apogee (terms for orbiting Earth, of course). At both points the motion is perpendicular to the applied gravitational force. So, the curvature is given from Newtonian mechanics as $f/(mv^2)$ where $f$ is the magnitude of the ...


10

If you only know two points on the orbit, and the mass and position of the central body, all you can determine for certain is its orbital plane. An example of two very different orbits that pass through a chosen $P_1$ and $P_2$ More geometrically: Given a focus $F_0$ and any two points $P_1$ and $P_2$, you can define a hyperbola with foci at $P_1$ and $...


10

Captain Wally Schirra was the first person to ever successfully pull off a space rendezvous. Here is, more or less, how he would have answered the question (this is a quote from Capt. Schirra after Gemini 6A): "Somebody said ... when you come to within three miles (5 km), you've rendezvoused. If anybody thinks they've pulled a rendezvous off at three ...


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