# Tag Info

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.

0

Approximations are made: Earth's orbital velocity remains the same: Angle between the Earth and the Earth's perihelion $\theta$ is increasing constantly. Eccentricity is small enough, that ellipse can be approximated to be $r=a(1-e \cos\theta)$. Earth is at its perihelion on 4th of January, and its eccentricity is 0.0167, so the given formula can be ...

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 ...

0

If it's a hyperbola, the semi-vertical angle is half the angle between the asymptotes and the cutting angle of the cone is parallel to the axis of the cone. Otherwise, it's not unique. There's an infinite number of pairs of angles that would yield the same conic section. Edit: Nevermind. You can get hyperbolas at more than one set of angles too.

0

You can't prescribe the motion in the general case. You need to build a simulation. That is calculate the acceleration of each body at every instant and integrate over a small time period to find the change in position and velocity. Then re-calculate the accelerations given the new positions (and velocities) and repeat for many small time steps.

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 ...

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

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 ...

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 ...

0

When two object orbit each other, the shape of their orbit is independent of their relative mass. In fact they will each have the same shape of orbit, but scaled by the mass of the other. So if you have two objects of mass $m$ and $2m$ respectively, then the former will have an orbit (circle or ellipse) that is exactly twice as big as that of the other. ...

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 ...

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 ...

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 ...

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 ...

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 ...

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, ...

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

Figure out a planet in orbit of 360 degree. You would have a 45 degree axis at start and -45 degree axis at mid course of this planet around the orbit. It would means that this planet would have taken a speed or inclined throughout that course implying that from Her own, it developed a way to tilt like around an ice cream cone. This could means that a ...

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 ...

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 ...

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). ...

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 ...

0

To first order, no. The definition of an orbit is that it's a free-fall trajectory — since everything around you is always experiencing the same acceleration as you are, you cannot actually perceive this acceleration without some external reference (like measuring your velocity compared to the Earth, and how fast it changes). That said, the closer ...

-1

When you are in a synchronous circular orbit, the gravitational force equals the radial force, at all times. If the orbit is elliptical, the variations in the orbit will translate into force variations. Whether these variations would be perceived by humans, depends on how "elliptical" the orbit is, and how close to other celestial bodies the orbit takes ...

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 ...

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 ...

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 ...

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 $g$.

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.

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.

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 ...

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