When does Phobos collide with Mars?

[PMNIMG]

What is wrong with what?

By definition, $G$ is the constant that appears in the equation: $F=G\frac{m_1m_2}{r^2}$ where one has two point-like (or spherically symmetric) masses, $m_1$ and $m_2$ whose centers are separated by a distance $r$ and $F$ is the gravitational attraction between them.

you are right. But just for once follow my equation and calculate. It will be Earth's gravitational constant

 

[PMNIMG]

sorry

PeroK​

but if you have time, can you email me about details?

 

[Steve4Physics]

Oh, sorry. I think it's b

You need to be certain which one it is - they are very different questions!

 

[PMNIMG]

Yes. I am certain.
But just to be clear. what exactly does (a) mean?

 

[Steve4Physics]

Yes. I am certain.
But just to be clear. what exactly does (a) mean?

Option a) means this...

Suppose you are hovering, stationary (relative to the surface of Mars), 6000km above the surface of Mars. You drop a stone. How long does it take for the stone to hit the surface of Mars?

But you say that's not the question uou are asking.

Edited.

 

[PMNIMG]

What is diffrent bitween a and b...?

 

[Steve4Physics]

What is diffrent bitween a and b...?

I've just explained a) in Post #40. Please describe what you think b) means, in your own words.

 

[PMNIMG]

I think (b) means time for an object with the mass of Phobos to collide with Mars.

 

[Steve4Physics]

I think (b) means time for an object with the mass of Phobos to collide with Mars.

What are the initial position and velocity of this object?

 

[PMNIMG]

Isn't it linear velocity of Phobos and semimajor axis of Phobos?
then the velocity is the diffrence?

 

[Steve4Physics]

Isn't it linear velocity of Phobos and semimajor axis of Phobos?
then the velocity is the diffrence?

Then considering your 'object' would be no different to considering Phobos itself!

That's exactly the same as asking how long Phobos, in its current orbit, will take to hit the surface of Mars.

Sorry, I can't help anymore. Good luck.

 

[jbriggs444]

you are right. But just for once follow my equation and calculate. It will be Earth's gravitational constant

When you say "my equation", which of the many equations you have posted do you mean?

At a guess, you mean:

I told you G=4*pi^(2)/Mk

Here I expect that $G$ is Newton's universal gravitational constant, $M$ will be the mass of the Earth and $k$ will be the constant in Kepler's third law as it applies to the orbital period and orbital diameter of the moon.

Note that to some extent, this calculation is backward. We calculate $GM_{\text{earth}}$ based on the observed gravitational acceleration $g$ at the surface of the earth and the known radius of the earth. We measure $G$ with something like the Cavendish experiment and we infer $M_{\text{earth}}$ from there. But you are asking that I compute $G$ based on the mass of the Earth when the truth is that we compute the mass of the Earth based on $G$.

In any case, you asked for the calculation. I will show my work. We want to evaluate $G$ in $G = \frac{4 \pi^2}{Mk}$ where $k$ is the constant of proportionality in Kepler's third law for the orbit of Earth's moon.

We have $M_\text{earth} = 5.972 \times 10^{24}\text{ kg}$
We need $k$ from Kepler's third law: $T^2 = ka^3$
We solve for $k$ yielding $k = \frac{T^2}{a^3}$.
We substitute in $T$ which is about 2.4 million seconds (one sidereal month).
We substitute in $a$ which is about 384 million meters. (384 thousand km - orbital radius)
So $k = \frac{(2.4 \times 10^6)^2}{(3.84 \times 10^8)^3} = 1.02 \times 10^{-13} \text{ s}^2/\text{m}^3$
We now evaluate $G = \frac{4 \pi^2}{Mk} = \frac{4 \pi^2}{(5.972 \times 10^{24})(1.02 \times 10^{-13})} = 6.5 \times 10^{-11} \text{ m}^3/\text{kg}/\text{s}^2$

Which is roughly correct.

(I cheated and simply assumed that the units came out right without carefully checking. But on review, it looks like the units do indeed come out properly).

Note that because I showed my work, you can easily see whether I solved the wrong problem, used the wrong input data, fumbled the algebra, fumbled the calculation, used an equation incorrectly, or used the wrong units. You can point to a specific difficulty. You are not forced to say only "that does not match what I got".

 

[Mister T]

I wrote my working and that is all I did.

If you look at the worked examples in a textbook you will see many examples of how to show your work. Typically they start with an explanation of their strategy, write equations, substitute values into those equations, and solve for solutions. Words of further explanation are included in between the above steps when necessary.

Even if your instructor doesn't require you to show your work on your assignments or exams, you will encounter instructors in your future studies who do.

It's a skill you need to practice to be successful in your education and career.

 

[Mister T]

Isn't it linear velocity of Phobos and semimajor axis of Phobos?
then the velocity is the diffrence?

The semimajor axis is a distance. You can't subtract a distance from a velocity.

I think (b) means time for an object with the mass of Phobos to collide with Mars

All objects free fall at the same rate regardless of their mass.

 

[PMNIMG]

g is not constant.

 

[jbriggs444]

g is not constant.

$g$, the gravitational acceleration of Mars at a location somewhere between the surface of the planet and the orbital radius of Phobos is indeed not constant. It depends on the location. [It also depends on the frame of reference one adopts, but we can agree to use an inertial frame in which Mars is at rest]

$g$, the [apparent] gravitational acceleration of Earth experienced at a point at rest on the Earth's surface is more or less constant. There is a variation of about 0.5% depending primarily on the latitude and altitude of the point on the surface that one chooses, but physics textbooks traditionally assume a constant such as 10 m/s², 9.8 m/s² or 9.80665 m/s².

$G$, Newton's universal gravitational constant is, of course, something else entirely. It really is a constant as far as we can measure. It is usually assumed to be constant everywhere as part of the cosmological principle.

 

[haruspex]

g is not constant.

Sure, but what is that in reply to?
If it is in reply to

All objects free fall at the same rate regardless of their mass.

then you misunderstood the remark.
For a given set of circumstances (mass and distance of body creating gravitational field) all bodies in that circumstance fall with the same acceleration. The time to impact will also depend on initial velocity. So in your earlier post

I think (b) means time for an object with the mass of Phobos to collide with Mars.

the mass of Phobos is irrelevant.

The semimajor axis is a distance. You can't subtract a distance from a velocity.

I think @PMNIMG meant that the velocity of Phobos is what makes the difference between options a and b in post #31.

Earth's gravitational constant

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, $G\frac{M_{Earth}}{R_{Earth}^2}$.

g=-4*pi^(2)/r^(2)k
View attachment 341511
=2.908359*10^16, ...1
(12745.5=r_phobos, 3365.5=r_mars)
and I calculated square root(there was erratum in my first posting. It's 170539114m/s)of 1, then I have 170539114m/s

Contrary to your denial in post #3, that is effectively using conservation of energy, $\frac 12mv_f^2-\frac 12mv_i^2=GMm(\frac 1{r_f}-\frac 1{r_i})$.
But as I posted, that does not help you find the time.

 

[PeterDonis]

Homework Statement: Use the gravitational acceleration equation with distance to find the time the phobos fall on Mars
Relevant Equations: F=ma=GMm/(r)^2
a=g=-1.33/(r)^2

I tried to calculate it, but I think I'm going wrong way.
I found m^2/s^2 in the definite integral...
(170539114.487m^2/s^2)
I don't know what it means!!

We don't know what you mean either. Several people have tried to find out but you have not given a clear answer.

You have said that you don't mean this: how long would it take an object at the altitude of Phobos but with zero tangential velocity, to fall radially to the surface of Mars. But the equations you list are only useful if that's the problem you want to solve, at least as far as actually hitting Mars is concerned (see below).

You have implied that you mean something like this: how long would it take Phobos itself, in its current orbit, to hit the surface of Mars. But Phobos in its current orbit does not hit the surface of Mars at all. It's a stable Keplerian orbit. The equations you list can be used to show the details of the orbit of Phobos given its orbital parameters, but that very solution, as I've just said, does not hit Mars, so it's useless for answering the kind of question you appear to be asking.

So we're at a standstill here until you can clarify exactly what question you want the answer to, and if it's not one of the above, you will need to give new equations and a new attempt at a solution, since nothing posted so far appears to be relevant.

 

[jbriggs444]

If the intended question is about the orbital decay of Phobos then there are two obvious approaches to the calculation.

1. Observation. Note that Phobos currently decreases its orbital radius by about 2 meters (6 feet) per year. Do an extrapolation and *voila*.

2. Postulate a mechanism for orbital energy loss, e.g. tidal friction. Phobos orbits faster than Mars rotates. Compute how much the shape of Mars could be distorted by the presence of the orbitting satellite, postulate a lag angle for how far the bulges lag behind the position of phobos (due to the planets resistance to deformation). Compute the net torque resulting from the placement of the bulges. Compute the energy loss per year.

Compare the results to observation. Use this to calibrate a more refined projection for how the energy loss will vary once the orbital radius has changed significantly.

This sounds pretty challenging.

Optional: Try to decide whether the Roche limit will intrude.

 

[PeterDonis]

If the intended question is about the orbital decay of Phobos...

Then none of the equations posted so far are relevant.

 

[haruspex]

If the intended question is about the orbital decay of Phobos then there are two obvious approaches to the calculation.

1. Observation. Note that Phobos currently decreases its orbital radius by about 2 meters (6 feet) per year. Do an extrapolation and *voila*.

2. Postulate a mechanism for orbital energy loss, e.g. tidal friction. Phobos orbits faster than Mars rotates. Compute how much the shape of Mars could be distorted by the presence of the orbitting satellite, postulate a lag angle for how far the bulges lag behind the position of phobos (due to the planets resistance to deformation). Compute the net torque resulting from the placement of the bulges. Compute the energy loss per year.

Compare the results to observation. Use this to calibrate a more refined projection for how the energy loss will vary once the orbital radius has changed significantly.

This sounds pretty challenging.

Optional: Try to decide whether the Roche limit will intrude.

I tend to think option (a) is intended, but if option (b) I would combine those two methods:
Construct a model for how the orbital decay should vary as a function of orbital radius, then use the current decay rate to fix the parameter.

Edit: just thought of option c, the mythological reading: Phobos collides with Mars when Achilles gets shell shock.

 

[Mister T]

or is there another way?

See the 4th paragraph of Post #10.

 

[haruspex]

The idea is that the free fall is effectively the same as for a highly eliptic orbit, where the semi-major axis is the distance from which the object falls. Then the free-fall time is one quarter of the period of the equivalent orbit.

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.

 

[PeterDonis]

The idea is that the free fall is effectively the same as for a highly eliptic orbit, where the semi-major axis is the distance from which the object falls. Then the free-fall time is one quarter of the period of the equivalent orbit.

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

The comparison I have seen is between strict radial free-fall and a circular orbit at the altitude at which the free fall starts. If we idealize the scenario so that the free-falling object can pass through the planet without being obstructed, then the free-falling object will execute simple harmonic motion through the center of the planet and out the other side, coming to rest at its starting altitude but 180 degrees away, and then reverse that to return to its starting point.

The period of that simple harmonic motion (i.e., the time to return to the starting point) will then be the same as the orbital period of the circular orbit--and so the free-fall time to the center of the planet will be 1/4 of that. But that comparison does not treat the free-fall motion as a highly elongated elliptical orbit with the planet's center at one focus.

 

[Vanadium 50]

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

 

[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?