97

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 ...


74

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 ...


71

Maybe some visual intuition for what actually happens in the Hohmann transfer helps? It's already very close to what you are describing. In the top arc, the spacecraft (yellow), is going a bit slower than Mars (red), so it's indeed "waiting" for the planet to catch up to it. It only touches the orbit of Mars in a point, but that's all we need if we time ...


62

Going directly to the Moon would require a very small launch window. The Earth orbit before enabled a launch window of about 3 to 4 hours, see this question. Abort from an Earth orbit was possible when the second ignition of the third stage of the Saturn V failed using the Service Module engine to initiate a reentry. Time in orbit was used to complete the ...


56

Yes. 1st scenario: A spacecraft orbiting the Sun at Earth distance vs. Pluto distance, shedding its orbital velocity The orbital velocity decreases with distance, according to the following formula, where $r$ is the orbital radius, and $\mu$ is the mass parameter (it's just a shorthand we use) $$v_{circular} = \sqrt{\frac{\mu}{r}}$$ The orbital velocity ...


44

There is very little to gain by going straight to the Moon, and as @Uwe has said, it makes the timing of the launch extremely demanding. Let me have my go at explaining why there is very little to gain. The most fuel efficient way for a rocket to get from Earth to the Moon is basically to accelerate as close to the Earth as possible until it is moving at ...


43

Wouldn't i inevitably spiral to sun surface even if i was faster than 0km/s ? No. On reasonable timescales, an orbit will have a fixed distance of closest approach, called "periapsis." (These timescales shorten if you're close enough to what you're orbiting that an atmosphere can drag you down). You don't really need to "drop in straight line" (which ...


40

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 ...


38

It depends whether or not you want to orbit or land softly upon Mars, or just hit it. For the former, you have to match orbits with it, that probably means burning more fuel. For the latter, you can skip the Mars orbit injection and just crash. This is quite fuel efficient, especially as the reduced delta-V requirements mean that you need less fuel for the ...


38

Primarily, locations of spaceports would change. California, not Florida would host the NASA's main launch site. Russia would be in slightly better position, able to send rockets over the Black Sea, nicer inclinations than currently available from Baikonur - although Vostochny wouldn't happen or would be closer to Chita. ESA could forget about French Guiana, ...


37

You've hit on a really interesting question. To answer this, I'm going to look at JPL Horizons, using the center of the Earth and the center of the Moon as the distances provided. I'm going to look at each of the Apollo missions, with the time that they were orbiting the Moon, showing the max distance, with 10 minute increments included. All distances in kms,...


36

First of all, it can require a lot of computer power to compute trajectories if they involve multiple gravitational slinghots to reduce fuel usage. This isn't because computing each segment is hard but because the search space is at least potentially exponential in the number of gravitational slingshots. I am pretty sure that this is what the thing in The ...


35

To answer your title question: By using its engines. However you seems to be quite puzzled by the fact that velocity of an object can decrease and increase over the course of an orbit. If the orbit is perfectly circular, the speed will always remain the same (until thrusters are used). However, as is the case with Chandrayaan-2, most orbits are ...


34

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 ...


32

I think you may have a misunderstanding that isn't addressed by any of the other answers. It is true that most of a rocket's work in entering orbit is building up enough speed to reach orbital velocity. But you have to build up even more speed to make it to the moon. In fact, while they were on their way to the moon they were still in orbit around the earth,...


29

So let's break down the answer into bite sized chunks... How do you transfer a spacecraft from one solar system body to another? There are two main things you need to do. Set up an orbit that intercepts the orbit of your target planet/moon. Time it so that your spacecraft intercepts the target orbit at the same time as the planet/moon you're trying to ...


27

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 ...


26

tl;dr: I don't think there is any scenario where you can strike the Moon with low velocity by using a small impulse to leave orbit. You can hit sideways with an orbital velocity of about 1680 m/s, or vertically with escape velocity the square root of 2 larger at 2376 m/s. Let's say I launched something into lunar orbit with minimal of propellant - just ...


22

This question seems to hinge on a fundamental misunderstanding about space, that is, to be fair, extremely common among the general public. It's the idea that space has no gravity, so things in space are weightless. "But wait!" you say. "I've seen videos of astronauts in space, and they sure seem weightless to me." And you'd be right, they do seem ...


22

I believe the answer is yes, but just barely. The distance from the Earth to the moon varies significantly over time, from 356,400 to 406,700 km. I plugged the dates of orbital entry and departure for each of the lunar Apollo missions (8, 10-17) into pyephem to find the ranges of lunar distance. At Apollo 13's flyby, the moon was one day past apogee and ...


20

The center of the Earth is, for any reasonable approximation, in one of the focus points of an elliptical orbit. For a circular orbit, there is only one focus point, so the center of the Earth is in the center of the orbit. The plane of the orbit thus would intersect both the center of the Earth as well as the launching site. If the launch site was on the ...


20

Does pork-chop plots exist for Earth-Moon System? No, because the concept doesn't make much sense. The Earth-Mars configuration and Mars' eccentricity makes the cost to send a vehicle from Earth to Mars vary by a huge amount. Pork chop plots are a useful way to visualize these huge variations. The cost of sending a vehicle to the Moon on the other hand ...


19

Because the lunar landings happened at some latitude, the landing sites were subject to longitudinal drift due to the Moon's rotation around its own axis. Due to the small latitude of the first landing[1], less than 1 degree for Apollo 11, and the short stay on the Moon, The LM lunar landing site did not drift far from the CSM orbital plane. Hence, Apollo ...


19

Why zero excess velocity? Well, with almost zero excess velocity you can stay near Earth, but not too near. For example, the Spitzer space telescope did this to communicate with Earth while avoiding radiant heat from Earth. It's been drifting away, but slowly enough that other factors first reduced its effectiveness.


19

In a two body system the total energy, the sum of kinetic energy and potential energy is constant for each body. If the total energy is constant for each body, the total energy for the whole system is also constant. So energy conservation is valid for each body alone as well as the system. In a three body system energy may be exchanged between the bodys. ...


18

All that computing power is not dedicated to the Artemis project. As you quote in the body of your question, The new supercomputer will be used by more than 1,500 scientists and engineers from across the country, including on projects like developing a more efficient quadcopter or simulating the inside of our sun. Not all of this computing power is ...


18

Given the mass costs in terms of consumables and the risk and support costs of keeping humans in space for longer, it seems unlikely that the multiple Earth-Venus flybys used by a lot of robot probes to get out to Jupiter or in to Mercury will ever be a sensible choice for humans. A Jupiter flyby on the way to Saturn is probably a no-brainer apart from maybe ...


18

You need below 2866 m/s of orbital velocity at 1 AU to crash into the Sun. You technically don't need to slow down exactly to 0 m/s relative to the Sun in order to crash into it. Let's calculate the approximate velocity required to graze the "surface" of the Sun. This is an excellent answer on how to calculate apoapsis and periapsis of an orbit. So first, ...


17

could we launch a vehicle in space for Earth, stop right on Mars trajectory... Yes, you could have a trajectory that came to a stop (briefly) in the path of Mars like how a ball thrown upwards stops (instantaneously) before falling down, except in this case you'd have to be moving directly away from the Sun on a straight-line trajectory with eccentricity = ...


Only top voted, non community-wiki answers of a minimum length are eligible