When does Phobos collide with Mars?

[haruspex]

the free-falling object will execute simple harmonic motion through the center of the planet

Only SHM while inside the planet surely?

 

[PeterDonis]

Only SHM while inside the planet surely?

Yes, what I described assumes a "surface satellite", i.e., the circular orbit is just above the planet's surface. For a fall from the height of Phobos (semi-major axis roughly 3 times the planet's radius) the math would be more complicated.

 

[SammyS]

Before we get into arguing about the answer, can we please wait until we find out what the question is?

Well, so much for that reasonable request you made !

 

[PMNIMG]

There is no such thing. G is a universal constant. "g" usually is a shorthand for the acceleration due to Earth's gravity at Earth's average surface, $G\frac{M_{Earth}}{R_{Earth}^2}$.

please see previous conversation

 

[PeterDonis]

Phobos is going to collide with Mars.

Yes, eventually that will happen due to tidal effects. But it is not at all clear that that's what you are asking about. You can't seem to give a consistent description of what you are asking about. See below.

The speed of Phobos is less than the first cosmic speed.

I have no idea what you mean by this.

And it is (a)

That's not what you said before. Here are the options:

a) a non-orbiting object, released from a point on @Phobos’s orbit, to hit the surface of Mars? Or...

b) the time for Phobos itself to hit the surface of Mars (which I understand is 30 - 50 million years).

To which you responded:

Oh, sorry. I think it's b

So which is it, a) or b)? If it's a), then your question has nothing to do with when the actual Phobos will actually hit Mars due to tidal effects. If it's b), then none of the equations posted so far are at all relevant.

Please clarify.

 

[PMNIMG]

sorry. I edited.
and did I use wrong english by first cosmic speed?
I thought it was minimum speed at which an object orbits a planet.
Then How will I approach this problem?
You said
then none of the equations posted so far are at all relevant.

 

[PMNIMG]

so, it is (b). About Phobos

 

[PeroK]

Is that right? I would have thought that in the limit Mars, the focus of the ellipse, is at the far end of the ellipse, so the fall distance is the whole of the major axis, and the fall time is half the orbital time.
Of course, that ignores the radius of Mars.

You're right. It's a half, not a quarter. Mars is at the focus, not the centre.

 

[PeroK]

I don't think this is correct as you state it.

You can check it out by doing the calculations. But, as @haruspex pointed out, the free fall time is half the period, not a quarter.

 

[PMNIMG]

You're right. It's a half, not a quarter. Mars is at the focus, not the centre.

Could you explain it with more details?

 

[PeroK]

Could you explain it with more details?

This thread is 70 posts long and you've made no progress. I'm sorry that I'm not able to help you.

 

[PMNIMG]

You said

I highly recommend using Kepler's third law. It gives you a quick way to get a good approximation. The answer will be the time to fall to the centre of a planet, and not to its surface. The difference should be small if the object falls from far enough away.

Please explain about this

 

[PeterDonis]

so, it is (b). About Phobos

In that case, every single post that has been made so far in this thread is irrelevant to your question. That includes all the posts you have asked people to explain. Kepler's Third Law is irrelevant. The free fall time from the altitude of Phobos is irrelevant. The equations you wrote in the OP are irrelevant.

The reason Phobos will eventually hit Mars as described in option b) is tidal effects. So you need to go find the relevant equations for tidal effects for a satellite like Phobos whose orbital period is less than the period of rotation of the planet it is orbiting. Once you do that, you can start a new homework help thread based on those equations.

In the meantime, this thread is closed.