New answers tagged

2

It can't be done. Unless it's really, really solid (and that's not likely!) trying to shove it that hard will break it up instead. You would need to use a bunch of bombs to move it more gently.


2

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 ...


6

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 $10^{16}kg$ ...


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 ...


5

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 ...


25

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 ...


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 ...


92

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 ...


72

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 ...


3

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 ...


2

Earth's sphere of influence has a radius of about 924000km. A highly eccentric orbit with a perigee at 100km altitude and an apogee at the SOI radius has a semimajor axis of 465239km. Throwing that into the vis-viva equation $v^2 = GM\left(\frac{2}{r} - \frac{1}{a} \right)$ where $G$ is the gravitational constant, $M$ the mass of earth, $r$ the orbital ...


-2

Earth's gravitational sphere of influence is not infinite That's your problem. The force of gravity does have an infinite range. There is no place in the universe where earth's gravity is not felt. As a result, it does not matter where you start, in order to escape earth, you need 11 km/s relative to the earth. If you first get into orbit at 100 km, then ...


2

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-...


2

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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