New answers tagged

3

Not only could it be done, but you could in fact get the hardware right now. Behold, the RB2000 rocket belt: We can work out the probable $\Delta_v$ of the rocket via the good old rocket equation: $$\Delta_v = g_0 \cdot I_{sp} \cdot \ln \left( \frac{M}{M_e} \right)$$ Where $g_0$ is a standard gravity, $M$ is the fully fuelled mass of the rocket and $M_e$ ...


2

Almost certainly, yes, although no such body has been identified. Normally to self-round a body needs to be far too big for a human to jump off. However, there's another possibility--a body that melted. Consider a very dirty sun-grazing comet. The ices burn off, but suppose it goes so close that the rocks themselves experience surface melting. (The pass ...


4

Lensing is caused by the ability of a mass to bend light, which angle can be expressed as: $$\theta = \frac{\mu}{rc^2}$$ Where $\mu$ is the gravitational parameter, $r$ the distance to the mass, and $c$ the speed of light. From this, we can find the closest point where light bend around the body into a focus. $$r_{min} = \frac{r_{body}}{\sin\left(\frac{\...


8

Mimas is the smallest known self-rounding body, and we've already asked: Could a human jump off Mimas without return? The answer is no. But we all want to answer to be yes, so what if we drop the "jumping" requirement, and just ask if a human could escape a self-rounding body with only human power? The surface escape velocity of Mimas is around 159 m/s, ...


0

What everybody who is used to thinking about rockets takes for granted but which may not be intuitively clear is: Every gram of fuel burned to counteract gravity is a gram of fuel wasted. Think about the worst case: Hovering on top of a running rocket engine which is just strong enough to keep you afloat. You are burning fuel without going anywhere. Your ...


3

Phobos. If you agree to strech the "self-rounding" part enough to include it. Because of its rather complex form and composition, there are points over its surface where the escape velocity is below the average human's running speed. At Deimos (even less round, but still...) you even don't have to look for a special place.


4

The term you are looking for is most likely Lagrangian point. Often synonymous, but sometimes used less strictly is Libration point. It's important to be careful about exactly what we mean when we say "cancels out", "equilibrium" or "static". When we have more than one body, they can't all be static. If they didn't move relatively to each other, their ...


37

No. Saturn's moon Mimas is the smallest body in the solar system known to be rounded through self-gravitation, and it still has a surface escape velocity of 159 m/s, far above the speed achievable by the best human athletes.


2

Update: This is not direct answer to the question as asked (thanks to @ruakh for pointing this out in the comments), but rather a supplementary answer that outlines another gravitational factor that would affect motion of an object at rest inside ISS. According to This NASA article, an object would move inside ISS due to the mutual gravitational force ...


8

Tidal forces come from gradients in gravitational fields. Along the horizontal axis of the station, the distance to Earth stays the same, so there's no gradient and therefore no tidal forces. To find the gradient along the vertical axis, we can simply use the derivative. $$\left(\frac{\mu}{r^2}\right)' = -\frac{2\mu}{r^3}$$ At the the orbital radius of ...


4

You're asking four questions, of which I can answer two: 1) Are integrators such as Runge-kutta integrating acceleration and velocity without gravitational contributions? Numerical integration is a science by itself, and there is no such thing as the Runge-Kutta integrator. RK are a family of integrators and parameters are optimised for specific ...


1

Gravity will not keep a species out of space, although it can make it incredibly expensive. A resource-limited species might not be able to make it to space, though--I'm thinking of Jovians. Chemical rockets suffer the tyranny of the rocket equation, if you need more than 30km/sec to attain orbit I don't think you're doing it, period. However, that's not ...


0

In Russell Borogove's answer, they assert "Linear increases in gravity require exponential increases in the size and expense of the rocket, so at some point it becomes impractical." That is the physics answer, but it's slightly different from an economics perspective. A more precise statement would be that if the only variables are payload and gravity, then ...


-2

If gravity is higher, the density of the atmosphere would be higher as well, and probably also to a greater altitude as well as less would have evaporated into space. So hydrogen or helium balloons would rise up more rapidly or lift more weight to possibly higher altitudes. Maybe this would be the main way of accessing space travel on this imaginary planet....


8

A (very high) upper limit is defined by the thrust to weight ratio of the first stage engine itself. The engine without a tank and a payload would not be able to lift off if the thrust is smaller than its weight measured under the high gravity. An engine build for such an extreme gravity would need more structal weight than at Earth's gravity. The ...


4

Note that you seem to be assuming chemical propulsion. Nuclear propulsion would work against even stronger gravity, but there are major safety problems.


17

As this article points out, rockets quickly get impractical. For example, at 10 times earth gravity, the rocket's mass is comparable to the planet's mass, so that's definitely some sort of limit! But who said we have to use rockets? Suppose we build a monorail completely encircling the planet at some convenient height $h$ above the ground, and accelerate ...


61

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 ...


5

There is a very nice article about the synchronized variation of the debris motion at GEO, available here. Authors give an example of the 'well-known GEO stable plane, which is a fixed point of the doubly-averaged differential equations (governing inclination and right ascension of the ascending node). Objects iniitialized at this equilibrium ...


3

Only have time for a quick rundown without numbers. Short explanation of equlibrium points Even at the GEO layer, the earth has gravitational differences and it not completely round. Land masses have a greater graviational pull than oceans. This means that, except for the stable equilibrium points (and unstable, on paper), a satellite in GEO will ...


4

This depends on how stable your rocket is. If your rocket is aerodynamically stable, meaning its center of pressure is behind its center of mass, the rocket will likely be turned to its velocity vector (zero angle of attack) by aerodynamics alone. A gravity turn is optimized for the least manual maneuvering possible. Any launch trajectory besides a perfect ...


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