# Tag Info

93

Life forms are made up from materials already present in Earth. Thus, increasing population would not alter the overall mass of the planet, and can't impact its orbit.

19

Mass is conserved (up to whatever small contribution nuclear decay has to the overall loss of mass). All biological matter is just created from materials from the environment (we do eat, right?). With exception of an occasional space probe, shooting stars and solar wind effects, the earth can be considered a closed system in terms of matter exchange. ...

8

As others have said, the mass of the Earth doesn't go up because we are eating food produced on Earth. Suppose we somehow were importing really large amounts of extra-terrestrial food, now what happens? We roast. Food contains a fair amount of carbon. Carbon from extra-terrestrial food is just as much a greenhouse problem as burning fossil fuels. ...

5

Plenty of others made it clear that babies aren't made out of matter from somewhere else. But if the earth's mass were to increase ten percent by magic, its orbit would not change. Its momentum—its tendency to fly off into space in a straight line—would increase by ten percent, and the counteracting force of the sun's gravity would increase by ten percent. ...

5

Here is the proof: Please refer to the wikipedia page on eccentric anomaly for a diagram and a couple of intermediate formulae. For an ellipse with the usual formula $x^2/a^2 + y^2/b^2=1$, it is the case that $\sin E = y/b$, and also by studying the figure on the wiki page you can see that $\sin (\pi-\nu) = \sin \nu = y/r$. Thus the two results you wish to ...

4

The scenario you suggest is of course hypothetical, but in all cases you must conserve angular momentum and mass/energy. So for example: If you have a way of removing mass from a star in such a way that the mass disappears outside the orbit of the planets (in astrophysics this is accomplished simply by mass loss - either the star has a wind that expels mass ...

3

If at any time the speed of the planet in the reference frame of the star exceeds the escape velocity $\sqrt{2GM_\star/r}$, where $M_\star$ is the mass of the star and $r$ is the distance from the star to the planet, it will escape in a hyperbolic trajectory (or straight line if $M_\star\rightarrow0$). As noted in the other answers, the result of the ...

3

This is known as the two-body problem of modeling the interactions of two bodies. More specifically, it is called the Kepler problem, as the objects interact via an inverse-square force - gravity. If we define some parameter $u$ as $$u\equiv\frac{1}{r}\tag{1}$$ where $r$ is the radius of the orbit at some angle $\theta$, then, using the Euler-Lagrange ...

3

How long time does it take before three planets achieve the same relative position? The answer is never, except for the case when their orbital periods can be expressed with low integers, like the 4:2:1 resonance of Io, Europa and Ganymede However, what you are asking about is when they are going to be in almost the same position again, a quazi-period. To ...

3

The picture of a black hole being like a huge vacuum is pretty misleading, not to say wrong. The gravitational field around the black hole is exactly the same as around any other object of the same mass. Like any other object, it's perfectly possible to orbit a black hole, just like we orbit the Sun and don't fall in. In fact, if you were to magically ...

2

Your questions shows some confusions about how orbits work. Kepler's three laws are useful in describing planetary orbits, but they don't quite get you the entire way into a direct way to calculate the position $\mathbf r(t)$ given a specified time $t$. Instead, you require a solution of Newton's equations of motion for the planet. There are two ways to do ...

2

Back of the envelope: mean thermal velocity of the molecules or atoms times mass loss rate. At room temperature that may be something on the order of 1000m/s*1e-9kg/s =1e-6N. For a 1000kg spacecraft this amounts to 1e-9m/s^2 acceleration and a position error after a year of about 500km. OK, if you want to land on Mars, maybe you want to correct for that, ...

2

Assuming that the twin star system is isolated, there is no external force acting on it so the center of mass should not move if the system was at rest initially. Centre of mass doesnot move Centre of mass moves although no external force is applied and the system was at rest initially

2

The motion of a comet can be understood in terms of a couple of principles. First - Inertia. Newton's first law states that an object will remain at rest or in uniform straight line motion unless acted upon by an external force In other words - assuming that "something" had given a comet a velocity, it will keep going unless something changes that. ...

2

All speed is relative. But an object that starts from rest at infinity will reach a velocity of about 11 km/s when it hits Earth, if Earth is the only thing pulling on it. At the same time, Earth is moving with an orbital speed of about 30 km/s. Their relative importance will depend on the direction from which the meteor is approaching - but on the whole ...

2

Meteors are essentially bits of rock that are independently in orbit around the Sun and which cross the Earth's orbit. If the Earth happens to be there at the same time then it will enter the Earth's atmosphere and we will see a meteor. The velocity of meteors is related to how fast they we going in their orbit around the Sun, combined with how fast the ...

2

Let's simplify the scenario slightly - imagine an object at the distance of the Moon, that has had its angular momentum slowed sufficiently that it will approach the Earth to a distance less than the radius of the Earth plus the radius of the Moon. Distance Moon-Earth ~ 400,000 km Radius Moon ~ 1,700 km Radius Earth ~ 6,300 km Mass Moon ...

1

Imagine that the wheel is stationary. A force is applied which accelerates the wheel horizontally. This add translational momentum to the wheel. Now, since the wheel does not slide, a frictional force is produced at point P which produces a torque on the wheel. The torque causes the wheel to start rolling, adding rotational momentum. Once the wheel is ...

1

I mean, if you believe what you've just written, $$\theta(t_2)-\theta(t_1)=\theta(t'_2)-\theta(t'_1)$$ Then I think you have already proven it, with a little more formal calculus. First rewrite that expression above as $\Delta \theta(t)=\Delta \theta(t')$, with the prime indicating the other interval. If $$t_2-t_1=t'_2-t'_1,$$ Then just set $\Delta ... 1 In order to arrive to this answer you have to assume that the size of the planet and star is negligible compared to the size of the orbit of the planet. When solving this question you have to understand what each symbol stands for:$T$stands for the orbital period of the planets orbit;$r$stands for the semi-major axis of its orbit;$GM$can be combined ... 1 A spinning space station does not provide artificial gravity away from Earth, it provides artificial gravity toward the center of the station. The first effect the spinning surface has on an occupant is to provide a tangential acceleration through friction. Then, because the tangential direction runs into the wall, the occupant is "thrown against the wall" ... 1 Your final question very much correlates with a famous thought experiment.If the Sun was suddenly removed the planet s will still continue to stay in orbit. For 8 minutes and 20 seconds. This is because the speed of the space time fabric or simply putting gravity travels at the speed of light. That is, the earth will be devoid of sunlight and will move ... 1 I understand the second law, although I have no idea how to create a formula that uses it. The most general form of this law is the law of conservation of angular momentum$\frac{d}{dt}L_\text{tot} = 0$; Kepler's law is an approximation which holds because the Sun is more massive than the planets (so you can regard it as approximately being located at ... 1 First of all you should note that the orbit of such a satellite is stable orbit which means if you deviate it from the exact value of$r=r_0$by small amount it will not go away and fall to the earth rather it will have a radial simple harmonic motion about$r=r_0$.This is because$r=r_0$corresponds to the minimum of effective potential in which the ... 1 As you are trying to simulate the movement of a planetary system you are making a dynamical simulation so you should not describe the orbits geometrically but strive to obtain them as a consequence of solving dynamical equations (numerically if necessary). If you solve the dynamical equations you will obtain the positions and velocities of the planets at ... 1 You have got your forces and potentials mixed up. If the force is given by: $$F(r) = \frac{GMm}{r^n}$$ then the potential is the integral of this: $$V(r) = -\frac{GMm}{(n-1)r^{n-1}}$$ In the question you link the force has an inverse cubic dependance,$n=3$, while the gravitational force has an inverse square dependance,$n=2\$. We get a circular ...

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