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98

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 ...


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

Previously posted comments are correct: in free space (assumed free of any other bodies' gravity fields) there is no way to convert the reaction wheels' angular motion to translational motion. There is one tongue-in-cheek way: throw a reaction wheel off the spacecraft in the direction opposite the direction of the desired delta-V! ;-) If you abandon the ...


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 ...


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 ...


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 ...


22

One reason large rockets are launched directly up is structural. Cylinders are strong under compression, stacking cylinders on top of each other means the weight is symmetric, you need less structural weight to hold it all up. Launch it straight up and make gentle changes in direction and the forces are equally distributed through the structure all the way ...


18

It's a matter of optimal trajectory - pitch maneuver/gravity turn which depends on characteristics of the rocket, the atmosphere, gravity etc. In particular, for rockets with lower initial thrust-to-weight ratio, the trajectory starts almost vertical; "rounding" the angle to perfectly vertical makes the launchpad infrastructure and preparation process easier;...


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, ...


16

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 ...


16

A single number is impossible to give, unless you exactly specify the kind of metal. So I'll answer in the general sense. The first question to answer is "How much energy does the sheet absorb?". This is given by the Inverse-Square Law: $$ I={\frac {P}{A}}={\frac {P}{4\pi r^{2}}}$$ For Mars, this works out to $~ 589\frac W {m^2}$ of energy. The ...


16

And note that if you want to hit the sun the cheaper (but slow!) way to do it is to head out. 12.32km/sec will take you to infinity, at infinity a burn of 0m/sec will kill your orbital velocity and you'll come straight in. Of course this will take infinite time, but even going only as far as Jupiter's orbit means you use less energy to drop your periapsis ...


15

2kW is not that much on Earth You've mentioned radiation and convection in your answer (you forgot conduction). Turns out the properties of Earth's atmosphere make conduction and convection way better than radiation for moving heat around. For an illustration, consider the size of a portable, 2kW, oil-filled radiator: this one lists the size as ...


14

The eccentricity is 1.0. The eccentricity $e$ of an orbit can be found from the radius of apoapse and periapse as: $$e=\frac{r_a-r_p}{r_a+r_p}$$ and the semimajor axis $a$ can as well, from: $$a=\frac{r_a+r_p}{2}$$ If you throw an object horizontally (velocity perpendicular to position vector) you will end up in a closed orbit if you throw at slower ...


14

I am Patrick Shober (the lead author of the study). Thanks so much for checking it out! If you check out Figure 10 in the paper, I have plotted the specific angular momentum in the Sun-centered frame. So this shows how the meteoroid (the rock) gained energy during the close encounter but then lost a fraction of it due to the atmospheric passage. This can ...


14

It still has the resistance of terrain against wheels (well, weaker than comparable terrain on Earth due to lower gravity - but then the terrain is pretty awful for driving), the same friction of bearings and so on - a car driven through loose sand on Earth will stop really fast due to the sand resistance, and not due to air. Now if instead of a lunar rover, ...


11

@OrganicMarble nailed it: ...it looks like it's the distance from the ecliptic plane. Yep, it's height above/below the ecliptic, a way to represent 3D in a 2D plot. At first I thought they might be thrust vectors like these but no, these are ballistic arcs. Instead I am 99.44% certain that these lines are use to indicate height above/below the plane of the ...


10

The energy efficiency isn't terribly useful when applied to rocket engines for space exploration. Using this definition of engine efficiency a cold gas thruster is more efficient than an ion engine$^{*}$ It's concerning the efficiency of with which the energy extracted from the fuel is converted to a force on the vehicle. It basically says if your exhaust is ...


8

The eccentricity of a radial orbit is $1$, regardless of its energy. This is a class of orbits where the type of orbit cannot be inferred from the eccentricity alone. With a "traditional" parabolic orbit of $e=1$, the angular momentum $L$ has a well defined value, but the semi-major axis $a$ is not defined. In the case of a vertical bounded free-...


8

While this animation is somewhat misleading as the wavelength line seems to be following the position of the exoplanet as it passes the star, I think this is an example of Exoplanet Spectroscopy. Simply put, scientists measure the light output from a star. When an exoplanet passes between the star and our observation point, the total light level drops. This ...


8

As others have pointed out, a civilization with zero knowledge of relativity could have carried out all space missions so far, and even constructed the GPS network, if they had simply added in various ad hoc corrections based on experience, without any understanding of the underlying physics. However, it's easy to come up with examples where if you failed to ...


8

Does it have three unequal principal moments of inertia... Does the ISS have three distinct principal moments of inertia? The answer is definitely "yes". Any object has a mass matrix, $M$. In three dimensions, this is a $3x3$ symmetric matrix. Because $M$ is symmetric, such a matrix can be diagonalised, so that $P^{-1} \cdot M \cdot P = \bar{M}$ is a ...


7

Simply to lift get Phobos out of Mars orbit you would need to increase its orbital velocity by a factor of $\sqrt{2}$ (this is generally true for any object in circular orbit). Phobos orbits at about 2.1 km/s (Wikipedia) relative to Mars, so this is a delta-V of $2.1 \times (\sqrt{2} -1)$ which is about $0.9 km/s$. It's mass, same source is about $10^{16}kg$ ...


7

No, it's a textbook case of conservation of the linear momentum vector in the absence of any external forces. Linear momentum of a system Sum(mv) is a conserved quantity even if individual parts are allowed to change their momentum vectors. Actually reaction wheels also conserve angular momentum of the total system (ship + wheel) as well! But that's Okay ...


7

There's no fundamental physical limitation, but there's certainly a practical one. The rocket equation is $\Delta V = V_{Exhaust} \ln(M_{Total}/M_{Dry})$. The exhaust velocity of a typical chemical rocket is around 2500 m/s to 4500 m/s, the exhaust velocity of a steam rocket is only 500 m/s or so. You need a delta-v of around 2 km/s to reach the Karman line....


6

Exactly at the center-of-mass of the Earth there is no net gravitational force. Gravity comes from mass, and all the mass of the Earth is evenly distributed around the center-of-mass (by definition), so all the forces cancel out. As you get deeper and deeper, the net gravitational force falls -- linearly if density is constant. In practice, of course, you ...


6

Consider NASA's Gravity Recovery and Interior Laboratory (GRAIL) mission. Not only would each probe's positions need to be precisely known, but the effects of lunar gravity at their locations. This pairing of position and gravitational measurement could qualify for Relativity being necessary for the success of the mission. https://www.nasa.gov/pdf/...


6

From Viking '75 Spacecraft Design and Test Summary Volume 1: Lander Design page 76


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