?How to derive the equation of motion F=ma, given this info

[Mark44]

for |r’(0)|=u , what is this , is it r? or gravity i.e should i sub r=0 or G=0?

It's hard for me to believe that you are asking these questions. We are 36 posts into this problem and you don't understand what the symbols in this problem mean. Go back to post #1 - it says what u is.

There's nothing in this problem that suggests that r = 0.
Also, why would you think that G = 0? G is a positive constant. Setting G to 0 is like asking if you should substitute 0 in for $\pi$.

But if a particle of mass M did move from a fixed point O, would that mean r’=0

There are two masses in this problem. The one at point O has a mass of M units (the units aren't given). The particle named P has a mass of m.

In this problem, mass M doesn't move, so it serves no purpose to assume that it does.

Apparently you don't understand what the symbols you are using mean.

What do you think each of these means (in words)?
r
r(t)
r(0)
r'(t)
r'(0)

When you answer these questions I'll give some suggestions for what you wrote in the attachment.

 

[cloud360]

It's hard for me to believe that you are asking these questions. We are 36 posts into this problem and you don't understand what the symbols in this problem mean. Go back to post #1 - it says what u is.

There's nothing in this problem that suggests that r = 0.
Also, why would you think that G = 0? G is a positive constant. Setting G to 0 is like asking if you should substitute 0 in for $\pi$.

There are two masses in this problem. The one at point O has a mass of M units (the units aren't given). The particle named P has a mass of m.

In this problem, mass M doesn't move, so it serves no purpose to assume that it does.

Apparently you don't understand what the symbols you are using mean.

What do you think each of these means (in words)?
r
r(t)
r(0)
r'(t)
r'(0)

When you answer these questions I'll give some suggestions for what you wrote in the attachment.

i know what they mean, i don't know how to use the initial condition |r'(0)|=u.

r is the position vector function.

r' means velocity,

r'(0) means velocity at t=0, right? but then how do i use the initial condition |r'(0)|=u,as my equation has no time or variable t, to sub t=0 into it !

i don't know where to sub 0 into, in the pic i have highlighted it red, the 0.

i also made an error, when i wrote

"for |r’(0)|=u , what is this , is it r?", i missed out a 0, should have written

"for |r’(0)|=u , what is this 0, is it r?"

but to conclude, can you confirm if my latest solution is correct, up to the last few lines. the way i used the initial condition was i chose G=0, that's how i interpreted |r'(0)|=u. which is wrong

 

[cloud360]

I wrote

"But if a particle of mass M did move from a fixed point O, would that mean r’=0"

i had a previous question were it said "a particle of mass M is released from a fixed point O". but did not know if i should interpret as r'(0)=0?

should i interpret as r'(0)=0

 

[Mark44]

i know what they mean, i don't know how to use the initial condition |r'(0)|=u.

r is the position vector function.

Yes. What I wrote as r and r(t) mean the same thing - the position at an arbitrary time.

r' means velocity,

And so does r'(t). To answer your question below, r'(0) means the velocity at time t = 0.

Since position, velocity, and acceleration are all vector quantities, if they're talking about speed, they mean the magnitude of the velocity, or |r'(t)|. The speed at time 0 is |r'(0)| = u.

r'(0) means velocity at t=0, right? but then how do i use the initial condition |r'(0)|=u,as my equation has no time or variable t, to sub t=0 into it !

Time is implied in the problem when they say that the particle is initially at a certain position. You can (and should) translate that wording into what it means, which is r(0) = a. You don't need to put in a bunch of verbiage that says "initially" suggests t = 0.

Similarly, when the problem says the the particle's speed is u, this means |r'(0)| = u. You don't have to explain any more than that.

i don't know where to sub 0 into, in the pic i have highlighted it red, the 0.

i also made an error, when i wrote

"for |r’(0)|=u , what is this , is it r?", i missed out a 0, should have written

"for |r’(0)|=u , what is this 0, is it r?"

No, r' is not a function of position r -- it is a function of time t. I explained above what r'(0) and |r'(0)| mean.

but to conclude, can you confirm if my latest solution is correct, up to the last few lines. the way i used the initial condition was i chose G=0, that's how i interpreted |r'(0)|=u. which is wrong

Yes. You cannot "choose" a value for G. Setting G to 0 makes no sense at all.

Additional comments in another post.

 

[Mark44]

is my solution correct now. also i have a question, if you will kindly answer.

[PLAIN]http://img231.imageshack.us/img231/1016/solf.gif

From the line where you have v dv/dr = -MGr^-2 I would do it exactly like I said in post #5, which I copied here.

In one of your lines where you're integrating, you have a indefinite integral on one side and a definite integral on the other side. DON'T DO THAT. Also, you don't need to explain each and every piddly step, such as when you move the constant outside the integral.

Going from the differential equation r'' = -MG/r², I get
$$v dv/dr = -MG r^{-2}$$
So
$$v dv = -MG r^{-2}dr$$

Integration gives us
$$\int v dv = -\int MG r^{-2}dr$$
$$\Rightarrow (1/2) v^{2} = + MG/r + C$$
Multiply both sides by 2:
$$v^2 = + 2MG/r + C'$$
where C' = 2C

At t = 0, v(0) = u and r(0) = a, so
u² = 2MG/a + C'
$$\Rightarrow C' = u^2 - 2MG/a$$

Therefore
$$v^2 = + 2MG/r + u^2 - 2MG/a$$
as required.

 

[cloud360]

just to confirm, i should not use integrals with limits, it it wrong to use integral with limits like i did?

 

[Mark44]

It's wrong to have an indefinite integral on one side and a definite integral on the other side, which is what you had.

The problem is easier, IMO, if you don't use limits of integration.

 

[cloud360]

It's wrong to have an indefinite integral on one side and a definite integral on the other side, which is what you had.

The problem is easier, IMO, if you don't use limits of integration.

someone told me to use integral with limits in the question below, and i did, is that wrong to

[PLAIN]http://img861.imageshack.us/img861/1264/sol6.gif

I really want to use integral with limits because i am used to it

is my solution to the main question correct, i subbed r=a using |r(0)=a|. or is it completely wrong because i used integrals with limits to represent displacement?

[PLAIN]http://img36.imageshack.us/img36/6577/sol3i.gif

 

[Mark44]

someone told me to use integral with limits in a different question.

This is a silly reason.

I really want to use integral with limits because i am used to it

Then you should use limits of integration on both sides, not just one.

What you have is not technically incorrect, but it looks sloppy.

You have introduced an error in the 4th line from the bottom. See if you can figure out what you did wrong.

 

[cloud360]

This is a silly reason.
Then you should use limits of integration on both sides, not just one.

What you have is not technically incorrect, but it looks sloppy.

You have introduced an error in the 4th line from the bottom. See if you can figure out what you did wrong.

i forgot to get rid of that 0??

so how do you use limits on integration on both sides, is it like i did below

[PLAIN]http://img9.imageshack.us/img9/4716/sol8.gif

 

[Mark44]

i forgot to get rid of that 0??

No. That's not it.

so how do you use limits on integration on both sides, is it like i did below

[PLAIN]http://img9.imageshack.us/img9/4716/sol8.gif[/QUOTE]

No. See post #40.

 

[cloud360]

No. That's not it.No. See post #40.

cant figure it out.

i subbed r=a, because of the initial condition r(0)=a. was that wrong?

 

[Mark44]

For your integrals, the limits of integration on the left integral are wrong.

The last four lines of your attachment are harder to follow than they need to be.

In your attachment, where you have
v² = 2MG/r - 2MG/a + k

after that, say this:

v(0) = u and r(0) = a, so the equation above becomes
u² = 2MG/a - 2MG/a + k = k

so k = u²
Therefore, v² = 2MG/r - 2MG/a + u²​

BTW, those bullet points you put in at each line are distracting.

 

[cloud360]

For your integrals, the limits of integration on the left integral are wrong.

thanks for your help, so all i need to say is "v(0) = u and r(0) = a, so the equation above becomes..."

also, what's wrong with the limit of integration on LHS, why is it not "0 to v" for the LHS integrand

(next time i will remove the bullet points from any attachments)

 

[Mark44]

I showed you a way to do it without the integration limits. If you insist on putting them in, you figure out what they need to be.