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141

You are correct, the centre of the Sun is not the Solar System's centre of gravity. A diagram (courtesy Wikimedia Commons), showing how the barycentre of the Solar System has changed over time. The Sun is affected by the gravity of all planets in the Solar System, but you are right, it is most affected by the two most massive ones; Jupiter and Saturn. ...


124

The force of gravity decreases with distance. It follows an inverse-square relationship... essential to know when you're grinding out the math, but not essential to a conceptual understanding. The fact that gravity decreases with distance means that at some distance, it will be negligible; an object sufficiently distant from Earth may be considered to have "...


95

Escape velocity reduces as you get further away from the Earth. If you proceed upwards at a constant speed of 1 mph (which as noted will require continuous thrust to counteract gravity), you will eventually reach a distance where the escape velocity is equal to 1 mph. Then, you will have reached escape velocity and are no longer gravitationally bound to ...


93

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

Objects in orbit are attracted to each other, it's just their mass is small enough that the force of gravity between them is infinitesimal. Gravitational acceleration is dependent on mass and distance. In a scenario where a 150 kg astronaut is 10 m from a 80,000 kg Space Shuttle, the astronaut would be pulled toward the Shuttle at 5.336e-8 m/second squared. ...


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 ...


67

You don't need a space probe. Or an aircraft. Or even a car. NIST has measured the predicted general relativity time dilation due to a change in altitude on Earth of one foot!


61

The problem isn't so much that humans cannot sustain high G forces for any extended length of time: The problem is that rockets cannot. If a rocket could sustain 1 g acceleration for a bit over a day, we could go to Mars in a bit over a day. It instead takes several months to get to Mars because the rockets used to get there only fire for a few minutes. The ...


60

To sum up the answers: the escape velocity is the velocity that, at a given distance, is sufficient to escape the gravitational field so that no additional energy (= acceleration) is needed. That is, if you are 26000 AU from Earth, you don't need any more fuel to counteract Earth's gravity, you just float away. However, when at Earth's surface, you will ...


60

There's no "bright line" at which space travel would become impossible; a slightly stronger gravitational pull would require bigger and more expensive rockets. Linear increases in gravity require exponential increases in the size and expense of the rocket, so at some point it becomes impractical1. At some point there's a theoretical barrier (no material ...


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 ...


52

Using the approximation $$\Delta \mathrm{center}=\frac{m_p}{m_\odot}\cdot\frac{\mathrm{dist}_p}{r_\odot}$$ and data from List of gravitationally rounded objects of the Solar System - Wikipedia: Name Distance Mass/[kg] Mass Distance ΔCenter /km /kg /sun mass /sun radius ---------------------------------------------...


51

The reason the Space Station is called a micro-g environment rather than a zero g environment is because the Space Station is rotating, because it's in low Earth orbit, and because it's big (for a spacecraft). The Space Station nominally rotates at the orbital rate so as to keep the nadir-pointing windows pointing downward. This alone means an accelerometer ...


51

No. The moon isn't that big but it isn't exactly small either. The moon's mass is 73,500,000,000,000,000,000,000kg, that's 73 sextillion, 500 quintillion kilograms. If we moved the whole of mount Everest from the earth to the moon (162 Trillion kg, which is completely unrealistic for us to do) then that would equate to an increase of 0.0000000022%, which ...


50

The "gravity of Mars" is not a number but rather a complex field. The most recent is remarkably detailed, made up to spherical harmonics degree and order 120, described by 29,512 coefficients: These maps are made using orbiters (three orbiters in this case), not landers. A lander/rover can give just one local gravitational acceleration and direction, which ...


50

In addition to specific probes like the one mentioned by called2voyage, the effect is significant enough that it affects everyday operations. For example, the GPS constellation needs regular clock corrections because the satellite hardware sits much higher up the gravity well than the ground hardware. The Wikipedia page for gravitational time dilation ...


50

What you're missing is some combination of the following: objects launched from Earth orbit are still in orbit around the Sun, objects in orbit don't need fuel to stay in orbit. All the planets stay in orbit around the sun because orbits are generally long term stable. A spacecraft flying from Earth to Mars follows a loop, like in the picture below. ...


49

The trajectory was not only "unhindered" - it was enhanced! Knowing mass of the planet you can calculate very precisely how the trajectory of a probe flying by will be affected. You modify the trajectory on arrival in such a way, that the departure trajectory will be exactly as desired. And due to some rather unintuitive physics caveats, you can make it so ...


43

It's complicated! Keep in mind the distinction between weight and mass. On the moon, weight is 1/6 what it is on Earth, but mass is the same. When you hold your arm straight out, you have to exert force equivalent to its weight to hold it in place. So on the moon, that force is much reduced. When you move your arm, you have to exert force proportional to ...


43

Ignoring the major point that human tolerance of G forces is not the limiting factor on space travel, plenty of thought has been made on how to counteract G forces, not least by 60s sci-fi writers. You can find more information than you ever wanted at Projectrho on this topic. The general gist: for lowish accelerations like 2 G, you don't need to do ...


39

They used the Lunar Landing Training Vehicle, as pictured below. This had a jet engine to provide 5/6 of the lift needed to hover the vehicle, plus rocket engines that simulated the LM's engines. With the jet running, the LLTV felt like it weighed 1/6 of its actual weight, so it came pretty close to simulating moon gravity.


38

Short answer, No different from Earth in floating. Buoyancy in water or any fluid is based on the weight of water displaced. Floating is based on the weight of the item displacing water. This is ultimately ends up in comparing densities. If the density of the displacing object is greater than the density of the fluid it will weigh more and sink, if it's ...


38

Yes, time dilation was experimentally confirmed by Gravity Probe A, launched by NASA on June 18, 1976. The clock rates of two masers (one on the probe and one on Earth) were compared, and it was found that the difference matched what was predicted with an accuracy of about 70 parts per million. To address your question of challenges in designing the ...


35

53 m/s is the approximate terminal velocity of a human skydiver. The terminal velocity of a 7-ton metal dart is quite a bit higher. Larger objects tend to be affected less by atmospheric drag than smaller ones, all other things being equal. Terminal velocity also increases with altitude because the air is thinner. Assuming 7000 kg mass, 3.5 m2 cross ...


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 ...


33

You are confusing velocity and acceleration. If you were to jump standing on the surface of the Earth you might experience 8 m/s which is 17 mph velocity upward, but the acceleration of gravity would act to retard your motion, slowing your velocity down. If you have a high enough velocity, the effect of (de) acceleration can not slow you down ...


32

No, there would be no measurable effect. But we can consider two things: force and mass. Let's imagine we planned very poorly, and always landed our ferry craft in the same Earth-moon orientation (so that by landing, the moon was always pushed "away" from its current direction of motion). The gravitational force between the Earth and moon (the force ...


28

Wikipedia gives $0.51 {km \over s}$ or $510 {m \over s}$ escape velocity, so, no, no leaving Ceres by jumping. Following my earlier calculations, an asteroid of the radius of Ceres would have orbital speed at near-surface orbit of about $336 {m\over s}$, which is way beyond jump strength of anyone as well. Gravitational acceleration on the Moon is $1.6249 {...


27

You are correct that Voyager did not change from above escape velocity to below escape velocity shortly after launch. The plot is misleading in that it is just not very accurate right there at 1 AU. The plot lines are kind of thick and a smidge off. Now that I look at it more closely, the escape velocity line in that plot is wrong in other places as well. ...


26

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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