# Tag Info

23

But can someone please explain why this is without using pure algebra? I will try without a single formula. In Newtonian gravity, the gravitational force on a particle is proportional to the particle's gravitational mass; the more gravitational mass, the more the gravitational force. In Newtonian mechanics, the acceleration of a particle, for a given ...

16

For simplicity, consider a perfectly circular orbit; the gravitational acceleration is always at a right angle to the velocity vector. This means that the speed cannot change despite the fact that there is constant acceleration. Note that for the speed to change, there must be a non-zero component of acceleration parallel (or anti-parallel) to the velocity ...

11

You certainly know that all things fall at the same rate regardless of their mass (neglecting friction). An orbiting body is not different from a falling body in that the only force acting on it is the gravity of the thing it orbits, so there is no reason its mass should influence its orbit.

9

Why does the mass of the orbitting object have no effect on its revolution at all? It does have an effect! However, the effect is immeasurably small if the orbiting object itself has a very small mass compared to the object it is orbiting. The most massive object we humans have put into orbit is the International Space Station, with a mass of 419.5 ...

6

$$1- 0.01672*\cos(0.9856*(\text{day}-4))$$ This is an approximate expression. Term by term, $1$ The mean distance between the Earth and the Sun is about one astronomical unit. $0.01672$ This is the eccentricity of the Earth's about about the Sun. $\cos$ This is of course the cosine function, but with argument in degrees rather than radians. $0.9856$ This ...

5

Orbit is free-fall around a body, so it is for the same reason a feather falls as fast as a bowling ball (in a vacuum). In free-fall, $F = mg$ And we know that, $F = ma$ So we can substitute, $ma = mg$ And divide by $m$, $a = g$ Thus, no matter what mass is, acceleration equals $g$.

4

Well, if you have $2{\pi}r$ and you know the time in which you traveled it, then you could find speed. Nothing prevents this.

4

You are no doubt familiar with the apocryphal experiment of Galileo showing that falling bodies fall at a rate independent of their "weight". [ We should really say mass.] An orbiting body is just a special kind of falling body, albeit one that manages to miss the ground due its sideways motion. Hence the term "free fall" as used regarding astronauts or ...

4

No. The shape of the orbit, i.e. how elliptical it is, does not depend on the relative masses of the two bodies. All objects in the solar system orbit around the centre of mass of the solar system. For obvious reasons, namely that the Sun contain far and away most of the mass of the solar system, the centre of mass of the solar system is quite close to the ...

3

To say that the orbit becomes more circular the greater the Sun's mass is not true. Instead, the eccentricity (i.e. how much the shape of an orbit varies from being circular) is governed by a couple of factors. If you have a planet orbiting about the Sun with a mass much less than that of the Sun, and you know the following for an instantaneous point in the ...

3

One has to realize that Kepler's laws are a mere approximation. The motion of planets around the sun is a two-body problem. In case of such two-body problems, both the bodies revolve around the center of mass. But it turns out that Sun is much much heavier than the planets. So the center of mass of the system is very close to the Sun and Hence it is a good ...

3

This is a standard numerical integration problem. The simplest solution is the Euler method. At each time-step you update the position $$x(t+\delta t) = x(t) + v(t)\delta t$$ and the velocity $$v(t+\delta t) = v(t) + F(x(t))\delta t/m$$ for each particle, where $F$ is the force acting on the particle and $m$ its mass. A more advanced method is Verlet ...

3

By the sounds of it you have made a mistake with the units. In fact, you should not be using SI units at all in your simulation; astronomical values in SI units vary by such huge orders of magnitude that they are often a source of floating point errors that can destroy trajectories. You should instead use the astronomical system of units. Specifically, ...

3

I like Kieran Hunt's answer but I'm going to give a different answer, even though I agree with what he said. In a very real sense, our solar system doesn't obey Kepler's laws because there are many bodies. The planets and even more so, the moons in our solar system don't precisely follow Kepler's 3 laws, but they mostly follow it pretty close. Our moon ...

3

If the rope is "radially directed" it means every point has the same angular velocity $\omega$. Assume a length $2\ell$, then you can integrate the force on the rope from $r-\ell$ to $r+\ell$ - gravitational force must equal centripetal force. This gives you an equation for $\ell$ as a function of $\omega$ and $r$. See if that gets you going.

3

The rotation period $T$ is given by $$T=2\pi \sqrt{\dfrac{a^3}{G(M_\text{Sun}+M_\text{planet})}}$$ where $a$ is the sum of the half axes of the ellipse. Routhly: $M_\text{Sun}=2\times 10^{30}$ kg $M_\text{Earth}=6\times 10^{24}$ kg $M_\text{Jupiter}=2\times 10^{27}$ kg If you assume both Earth and Jupiter are orbiting around the Sun (and neglect the ...

3

I was wondering that if they were orbiting in same orbit then will they both have same time period? If yes, then why because as they both have different angular momentum and both have so much of differences. I'll break this down into two parts, first looking at the period of individual objects orbiting the Sun at a distance of one astronomical unit (but ...

2

A collapsing gas cloud is an open system. It loses mass, energy and angular momentum as it collapses. Even if the net angular momentum of the cloud is zero, after the collapse the final planetary disk can have a significant net angular momentum, and the ejected material will have the opposite angular momentum. What can not happen, and that's where your ...

2

I had the same feeling as you when I watched the video again recently. It seemed like one of the ice giants would get ejected after coming too close to Jupiter. It turns out that there's a name for this: the jumping Jupiter scenario. Outside Wikipedia, it's described in Fassett & Minton (2013) (paywall!) and tangentially in Deienno & Nesvorny (2014). ...

2

Let's consider a hover craft or a rocket that will accelerate up to a certain elevation and keep hovering at it, as a better vehicle for our purposes than a plane. Let's forget about air momentarily. What will happen is that the moment our air craft leaves contact with the ground, its initial tangential velocity will be the same tangential velocity of ...

2

For a conventional planet (i.e. one that is self gravitationally round), conservation of the planet's angular momentum makes this impossible (except for the trivial case of the axis perpendicular to the orbital plane. A non-spherical planet with a tilted axis will precess under the influence of tidal forces. It takes the earth about 26,000 years to go ...

2

Yes, objects with mass all attract to each other and move each other of course, except that the star doesnt change it's theoretical orbital shape depending on it's mass, it probably just experiences small tidal forces that aren't much bigger than it's own centrifugal forces. The oscillation of the star position is a complex dynamic based on it's surrounding ...

1

Yes, both special relativity and general relativity have to be taken into account. The total time dilation is given by $$\frac{\text{d}\tau}{\text{d}t} = \sqrt{ \left(1 - \frac{2GM}{rc^2}\right) - \left(1 - \frac{2GM}{rc^2}\right)^{-1}\frac{v^2}{c^2}},$$ where $\text{d}\tau$ is the time measured by a moving clock at radius $r$, and $\text{d}t$ is the ...

1

The International Space Station is in low Earth orbit, only about 400 km above the surface according to this page, or about 1.063 time Earth's radius. This means that, air resistance aside, getting the International Space Station to escape the Earth's gravitational pull would be nearly as difficult as launching a payload with the same mass from the Earth's ...

1

The answer based on accepted physics is trivial: none. That's why they call it science FICTION. One can theoretically reduce the amount of propellant, but then the required power these ships would have to develop would go trough the roof. Keep in mind that a medium sized rocket develops on the order of 10GW of power for a few minutes during first stage burn ...

1

If I understand the question right, we suppose we want to prove to someone that the earth orbits the sun. I'm not quite sure that' the case from a scientific point of view. Literally speaking, we can choose any reference frame we like and thus prove a heliocentric system or a or a geocentric. Quoting Einstein:" The struggle, so violent in the early days ...

1

A presentation on the SETI Weekly Seminar series (available on Youtube) points out that tidal locking (e.g. expected of a planet in the habitable zone of a red dwarf) can involve higher muliples than same-face-shows, and in fact an eccentric orbit favors an odd half multiple (3:2 like Mercury). There is also orbital inclination to consider. The pattern of ...

1

Acceleration is any change in velocity, whether it's the direction of the velocity or its magnitude. In the case of a perfect orbit it is the direction, the tangential velocity of the satellite should keep a constant magnitude and just change direction as the satellite orbits. When the velocity is slightly less than what is needed for a perfect orbit, it ...

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