# Tag Info

1

Direct integration schemes give bad results. You can do much better by using the exact solution in the absence of gravitational interactions between the planets. You can then set up a variation of constants approach where you take the integration constants (which are the orbital parameters) as dynamical variables and write the differential equations (where ...

3

The standard way to choose a time-step is to run a test simulation with multiple bodies and plot the total energy of the system versus time. The total energy should remain (roughly) constant. If your step size is too large then you will get energy drift. So simply find the largest time-step that does not produce energy drift. In the case of modelling the ...

0

If the moon were not present the rate of precession would be much less than the current figure of 1.4 degrees per century. More seriously, the tilt of the earth's axis would vary over a much greater range than at present and this could have grave consequences for the climate which would show much larger variations than have been known up to now. The ...

0

As far as I understand your question, you can consider the Moon as continuously falling down to the Earth (or continuously following the moving Earth, if you want) due to the gravity. Nothing compensates gravity. However the problem is that the Moon aims not precisely into the Earth, but a bit away - that was original "error" from the Moon's first day. The ...

0

Lets distinguish 'rotation' into 'spin' (rotating around itself), and 'orbit' (orbiting around another object). The orbital motion is due to gravity. The 'centrifugal force' (instead of 'kinetic energy' of rotation, per se) is balanced by the 'gravitational force' [1]. The spin of an object doesn't need an external force (gravity from another object) to ...

2

Just to more fully answer your question, here is an example of what differences in distances can mean mean as far as they affect gravitational forces. The planet Jupiter is extremely massive, and one side of one of it's moons, Io, feels a slightly larger gravitational pull than the opposite side. This difference in distance results in a gravitational force ...

2

I think you are confusing the escape velocity with terminal velocity. While for the local velocity the limit is only the speed of light, terminal velocity is achieved much sooner because of the air resistance. The equation can be found here and depends on the shape, size and density of the asteroid. For particles travelling near the speed of light you have ...

2

For the sake of simplicity (at the expense of real-life accuracy), let's assume that an asteroid is already travelling at some speed $v_0$ directly toward the Earth, and it never deviates away from that direct line path (despite the revolution of the Earth around the Sun). Let's also assume the distance from Earth is very large compared to the radius of ...

-5

No, you won't speed up and the fastest speed the earth can accelerate you to is at escape velocity.

0

First, calculus isn't just really small steps: I can show you limit processes that disagree with any really small step based solution, like making a staircase with smaller and smaller treads. The total "tread plus rise" size remains 2k, while the limit is a line with length sqrt(2)k. However, almost all of the parts of calculus that work in predicting ...

4

Since you still seem puzzled I'll try a different tactic here: You're showing the tangential velocity (A) and the radial acceleration (B) and adding them to get the green arrow. What you're missing is that this occurs in a gravity field. As the initial path climbs away from the object it's orbiting it loses velocity. This shows up as a third arrow ...

3

After reading the answers from other people, I see that you are still confused, so I thougth i coult take my chances on clearing up your confusion. It seems that your confusion is why does the orbiting body not accelerate to infinite velocities (or crashes to body A) eventually, and the origin of your confusion is in your assumption that the resulting ...

5

Your vector sum drawing is incorrect. Considering the simple case of a perfectly circular orbit, there is a tangential velocity vector and a perpendicular (inward) acceleration vector. Strictly speaking, you can't add them because they are different quantities (acceleration vs velocity). At some point in the orbit, the object has an initial velocity V in ...

14

Because the direction of the velocity changes. The velocity will start to point less and less 'towards' point A and when the distance between A and B is the smallest, the velocity will make a right angle with the radius, which means acceleration vector also makes a right angle with the velocity. At this point the radial component of the speed is zero and the ...

12

This question points out the importance of symplecticity in physics. In an orbital simulation, suppose one simply advances state via \begin{align} \boldsymbol x(t+\!\Delta t) &= \boldsymbol x(t) + \boldsymbol v(t)\, \Delta t \tag 1 \\ \boldsymbol v(t+\Delta t) &= \boldsymbol v(t) + \boldsymbol a(t)\, \Delta t \end{align} where $\Delta t$ is a ...

20

You have to consider the limit of infinitesimally short time, in which the (vertical on the paper) component of velocity is infinitely short, and thus also the angle changes for an infinitesimal amount. In this limit, the correction to the length is quadratic in the time step and vanishes exactly in the physical limit of continuous time. Pythagoras: ...

5

Any body travelling with increasing velocity increases its kinetic energy $KE$. Since in your system: $$KE+PE=\text{constant}$$ where $PE$ is potential energy. Therefore increase in $KE$ results into decrease in distance between the objects (so as increase $PE$). Note: $KE$ is always positive. $PE$ can be positive or negative. $PE$ is negative for bound ...

1

It's worth thinking about why the tidal bulges1 are not lined up with the line between the bodies (which is where you would naively expect them to) and then thinking about how that affects the gravitational interaction between them. Because the moon takes about one month to orbit and the Earth takes one day to turn, the naive location of the tidal bulges ...

2

The angular momentum is conserved in central force motion (like what we have in the case of Earth-Moon system). In such a case, the force $\vec{F}$ and the radius vector $\vec{r}$ are parallel so that the resultant torque is zero $$\vec{\tau}=\vec{r}\times\vec{F}=o$$ This means the angular momentum ($\vec{L}$) of the of the Moon about the center is a ...

0

It is not clear at which level you want to simulate this. Fine grained If you want to do this on a very fine grained level, you just need so simulate Newtonian mechanics and gravity and it should emerge from itself when you have a trajectory set up. This means that you compute the gravitational force to be $$\vec F = - G \frac{Mm}{r^3} \, \vec r \,,$$ ...

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