When does Phobos collide with Mars?

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The discussion revolves around calculating the time it would take for Phobos to collide with Mars, with participants expressing confusion over the equations and methods used. The original poster struggles with integrating gravitational acceleration and understanding the implications of their calculations, which resulted in a velocity squared value rather than a time estimate. Several contributors emphasize the need for clarity in the equations presented, specifically pointing out the importance of using Newton's laws and Kepler's Third Law appropriately. There is a consensus that the problem requires a more precise formulation and a better understanding of the physics involved, particularly regarding the initial conditions of the object in question. Ultimately, the conversation highlights the complexity of the problem and the necessity for clear communication in mathematical problem-solving.
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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!!
 
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PMNIMG said:
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!!
Please post your working.
It sounds like you found the square of the impact speed by using energy conservation, but that won't give you the time.
 
I wrote my working and that is all I did.
and I didn't use energy conservation. I think I can find time by Newton's second law. My question is that.
How can I get the left side?
and If there is no answer in this problem, then please say so.
 
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The question is not entirely clear. I assume it means find the time that a body with no tangential velocity would take to all to the surface of Mars from the orbit of Phobos. Is that how you read it?
PMNIMG said:
I wrote my working
I see no working.
You wrote two equations giving acceleration as a function of distance, then came up with a value which, by its units, appears to be the square of a velocity.
How did you get from the former to the latter?
You also mention an integral, but did not post one.
PMNIMG said:
I think I can find time by Newton's second law.
Finding time when the acceleration is a function of distance is not trivial. You cannot use SUVAT kinematic equations as those are for constant acceleration. How do propose to do it?

PMNIMG said:
How can I get the left side?
Left side of what?
PMNIMG said:
and If there is no answer in this problem, then please say so.
There has to be an answer since the physical arrangement is fully specified. Whether there is a closed-form analytic solution is another matter.
 
My working is the eqution. I used Kepler's Third Law to find 1.33*10^20(I mis), and I didn't posted integral
because it's just calculating. I did a definite integral from Phobos to the center of Mars. what left side? equation of motion. I know SUVAT doesn't help. It's not acceleration I know.
 
PMNIMG said:
My working is the eqution. I used Kepler's Third Law to find 1.33*10^20(I mis), and I didn't posted integral
because it's just calculating. I did a definite integral from Phobos to the center of Mars. what left side? equation of motion. I know SUVAT doesn't help. It's not acceleration I know.
That still does not qualify as showing your work. Show us the integral. Show us its limits. Show us how you solved it. Tell us what you think it means. Show your units. Tell us how you used Kepler's third law, what input data you used, what output data you computed, what units were used and how that result is relevant.

You are coming here with a result that is in the wrong units. That means that you have computed it wrongly. In order to diagnose that failing, we need to see the work.
 
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OK. Then I'll write the calculation

Kepler's Third Law
T^2=ka^3
g=GM/r^2
G=4*pi^(2)/Mk
g=-4*pi^(2)/r^(2)k
1709985495120.png

=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
 
Maybe I did it all wrong and it could be meaningless value. But still. Please tell me how to approach this problem
 
PMNIMG said:
OK. Then I'll write the calculation

Kepler's Third Law
T^2=ka^3
g=GM/r^2
G=4*pi^(2)/Mk
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
More. We still need more. I struggle to read between the lines.

You wrote down four unmotivated equations. You did not solve any of them. You then wrote down some numbers in a definite integral. You did not motivate the integral. You did not tell us where the ##1.33 \times 10^{20}## came from. You did not show units on anything.
 
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PMNIMG said:
Maybe I did it all wrong and it could be meaningless value. But still. Please tell me how to approach this problem
Let me say first that your work is practically impossible to follow. You need to develop communication skills. It's not enough just to type in some numbers and expect others to understand what you are trying to do.

There are two ways to solve this problem. The first is to integrate the varying force/acceleration as the object falls. This requires some advanced mathematics. There are some threads on here (if you can find them) where a full solution is presented, including a calculation of the time to fall.

The second method involves the trick of using Kepler's third law and treating the fall as a "degenerate" orbit. It looks to me that you might be mixing up the two solutions.

In any case, let's focus on the method of using Kepler. 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.

Does that make sense?
 
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  • #11
jbriggs444 said:
More. We still need more. I struggle to read between the lines.

You wrote down four unmotivated equations. You did not solve any of them. You then wrote down some numbers in a definite integral. You did not motivate the integral. You did not tell us where the ##1.33 \times 10^{20}## came from. You did not show units on anything.
obviously 1.33*10^20 is 4*pi^(2)/k
and is Kepler's Third Law and Gravity acceleration formula unmotivated?
 
  • #12
PMNIMG said:
obviously 1.33*10^20 is 4*pi^(2)/k
and is Kepler's Third Law and Gravity acceleration formula unmotivated?
I definited integration of gravitational acceleration!
 
  • #13
PMNIMG said:
I definited integration of gravitational acceleration!
Your solution looks wrong. You would need first to set up an appropriate differential equation. Note that this will be a complicated solution with a lot more mathematics than you have presented.
 
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  • #14
what equation should I use?
 
  • #15
or is there another way?
 
  • #16
PMNIMG said:
obviously 1.33*10^20 is 4*pi^(2)/k
Thank you for explaining where one of your numbers came from. Units would be nice.

No, it was not obvious to me. If I were trying to integrate gravitational acceleration over distance (yielding work per unit mass) I would not have bothered with ##k##. I would have simply used Newton's universal gravitational constant ##G## and the mass of mars, ##M##. Both of those were available as inputs.

The result of that integral would be in units of energy per unit mass. How would that be useful to obtain time of transit? Unless one planned to use the figure to compute velocity as a function of radius and do a second integration of ##\frac{1}{v(r)} dr## to get cumulative elapsed time. (I think that is where @PeroK was going with the approach that he did not recommend).

PMNIMG said:
and is Kepler's Third Law and Gravity acceleration formula unmotivated?
Yes. Invoking the laws is not good enough. Explaining how those formulas fit the problem and what you are solving each equation for would provide the motivation.
 
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  • #17
I told you G=4*pi^(2)/Mk
 
  • #18
  • #19
4*pi^(2)/M_(earth)k.
and k is 2.97*10^(-11)
 
  • #20
What you're saying is earth's G
 
  • #21
PMNIMG said:
4*pi^(2)/M_(earth)k.
and k is 2.97*10^(-11)
Always remember to use units. In other units, k=1.
 
  • #22
PMNIMG said:
What you're saying is earth's G
No. ##G## is Newton's universal gravitational constant. It works everywhere.
 
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  • #23
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.
 
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  • #24
jbriggs444 said:
No. ##G## is Newton's universal gravitational constant. It works everywhere.
No. If you derive the law of universal gravitation, you can have your answer.
F=4*pi^(2)m/kr^2=GMm/r^(2)
 
  • #25
so G is not constant
 
  • #26
or is there anything wrong in my certification
 
  • #27
PMNIMG said:
so G is not constant
What are you babbling about now? ##G## is constant.

Edit: Sorry. I must learn to remain polite. My role should be as a mentor, not as a castigator.
 
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  • #28
Why? tall me what's wrong
 
  • #29
PMNIMG said:
Why? tall me what's wrong
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.
 
  • #30
PMNIMG said:
Why? tall me what's wrong
Whatever approach you take, you need to completely rethink this problem. I would start again. A precise statement of the problem would help.
 
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  • #31
@PMNIMG. You haven’t yet answered @haruspex ’s question in Post #4 (unless I missed it). Are you trying to find the time for:

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).

Which one? (Or is it something else?)

Also, it might help us if you say what course you are on.
 
  • #32
Steve4Physics said:
@PMNIMG. You haven’t yet answered @haruspex ’s question in Post #4 (unless I missed it). Are you trying to find the time for:

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).

Which one? (Or is it something else?)

Also, it might help us if you say what course you are on.
Oh, sorry. I think it's b
 
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  • #33
and if you want it explained theoretically, please derive the law of universal gravitation with Kepler's third law
 
  • #34
PMNIMG said:
you are right. But just for once follow my equation and calculate. It will be Earth's gravitational constant
Sorry, I'm out! Good luck with your studies.
 
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  • #35
  • #36
jbriggs444 said:
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
 
  • #37
sorry

PeroK

but if you have time, can you email me about details?
 
  • #38
PMNIMG said:
Oh, sorry. I think it's b
You need to be certain which one it is - they are very different questions!
 
  • #39
Yes. I am certain.
But just to be clear. what exactly does (a) mean?
 
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  • #40
PMNIMG said:
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.
 
  • #41
What is diffrent bitween a and b...?
 
  • #42
PMNIMG said:
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.
 
  • #43
I think (b) means time for an object with the mass of Phobos to collide with Mars.
 
  • #44
PMNIMG said:
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?
 
  • #45
Isn't it linear velocity of Phobos and semimajor axis of Phobos?
then the velocity is the diffrence?
 
  • #46
PMNIMG said:
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.
 
  • #47
PMNIMG said:
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:
PMNIMG said:
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".
 
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  • #48
PMNIMG said:
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.
 
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  • #49
PMNIMG said:
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.
PMNIMG said:
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.
 
  • #50
g is not constant.
 
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