When does Phobos collide with Mars?

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Homework Help Overview

The discussion revolves around calculating the time it would take for Phobos, a moon of Mars, to collide with the surface of Mars, using gravitational acceleration equations and concepts from classical mechanics. The participants are exploring the implications of their calculations and the relevant physics principles involved.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to use gravitational equations and Kepler's Third Law to find the time of fall. There are questions about the interpretation of results, the meaning of units, and the appropriateness of methods like energy conservation and Newton's second law in this context.

Discussion Status

The discussion is ongoing with various interpretations being explored. Some participants have provided guidance on the need for clearer communication of calculations and assumptions. There is recognition that the problem may not have a straightforward solution, and the complexity of the mathematics involved is acknowledged.

Contextual Notes

Participants are grappling with the definitions and assumptions related to gravitational acceleration and the setup of the problem. There are mentions of missing information and the need for more detailed workings to clarify the calculations presented.

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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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  • #10
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.
 
Last edited:
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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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