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

[cloud360]

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

Homework Statement

[PLAIN]http://img861.imageshack.us/img861/1557/2009b5a.gif

Homework Equations

The Attempt at a Solution

Is my attempted solution correct?
[PLAIN]http://img42.imageshack.us/img42/2465/2009b5asol.gif

 

[cloud360]

Also, i would like to know if my solution to part b is correct. I have provided question and solution below, can someone kindly tell me if my solution is correct
[PLAIN]http://img863.imageshack.us/img863/8066/2009b5bsol.gif

 

[Mark44]

Homework Statement

[PLAIN]http://img861.imageshack.us/img861/1557/2009b5a.gif

Homework Equations

The Attempt at a Solution

Is my attempted solution correct?
[PLAIN]http://img42.imageshack.us/img42/2465/2009b5asol.gif[/QUOTE]
No. At the bottom of what you wrote, you have F = GMm/r². Then you say that F = |GMm/r²|, and somehow turn that into -GMm/r².

First off, force F is a vector (has direction). It's not a magnitude, so you can't just stick absolute values in.

Particle P starts off at speed u, so we can assume that it is not being accelerated. When it's moving, the only force on it is the force to to the gravitation from mass M. What direction is P moving? What direction is the force of attraction to M?

What you are forgetting is that F = ma = mr''.

 

[cloud360]

No. At the bottom of what you wrote, you have F = GMm/r². Then you say that F = |GMm/r²|, and somehow turn that into -GMm/r².

First off, force F is a vector (has direction). It's not a magnitude, so you can't just stick absolute values in.

Particle P starts off at speed u, so we can assume that it is not being accelerated. When it's moving, the only force on it is the force to to the gravitation from mass M. What direction is P moving? What direction is the force of attraction to M?

What you are forgetting is that F = ma = mr''.

Do i have to remember the formulae F=G*(Ma)(Mb)/(Rab)^2, which i gave

because i found that fomrulae on wikipedia and don't know if that is a general true formulae .

also, someone told me a=r''=r''ȓ, where ȓ is a unit vector.

is that true, if so how can i use it, as i don't know how to get negative part

 

[Mark44]

For the b part, it looks like you have the right idea, but you could say it a lot more cleanly. For one thing, in an indefinite integral you need to include the constant of integration. For a definite integral, you don't need the constant of integration. It looks like you are adding the constant of integration when you do a definite 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.

 

[Mark44]

Do i have to remember the formulae F=G*(Ma)(Mb)/(Rab)^2, which i gave

I don't think so, but I suppose it depends more on what your instructor requires you to have memorized.

because i found that fomrulae on wikipedia and don't know if that is a general true formulae .

also, someone told me a=r''=r''ȓ, where ȓ is a unit vector.

is that true, if so how can i use it, as i don't know how to get negative part

I don't think the unit vector business is useful in this problem.

Just answer these questions:
In which direction is the particle moving?
In which direction is the force of attraction?

 

[cloud360]

I don't think so, but I suppose it depends more on what your instructor requires you to have memorized.

I don't think the unit vector business is useful in this problem.

Just answer these questions:
In which direction is the particle moving?
In which direction is the force of attraction?

It says that it's moving radially.but how should I interpret that

 

[Mark44]

It says that it's moving radially.but how should I interpret that

Look at the sketch you show in post #2. Take a radius of mass M and extend it outward. Particle P is moving along that line.

There's something else you have in post #2 is partially incorrect, and might be causing problems for you.

Anything distance = magnitude

.
There's the concept of directed distance that you are probably missing. On the real number line, +5 and -5 are both 5 units from the origin, but +5 is to the right of the origin and -5 is to the left of the origin. In all of the problems you have posted, it's important to keep track of the sign of the velocity. If v (= r'(t)) is positive, the particle is moving in the direction of positive r (usually this is to the right). If v is negative, the particle is moving in the direction of negative r.

Relative to the sketch you drew in post #2, in which direction is the particle moving?
In which direction is the force of attraction?

 

[cloud360]

Look at the sketch you show in post #2. Take a radius of mass M and extend it outward. Particle P is moving along that line.

There's something else you have in post #2 is partially incorrect, and might be causing problems for you.
.
There's the concept of directed distance that you are probably missing. On the real number line, +5 and -5 are both 5 units from the origin, but +5 is to the right of the origin and -5 is to the left of the origin. In all of the problems you have posted, it's important to keep track of the sign of the velocity. If v (= r'(t)) is positive, the particle is moving in the direction of positive r (usually this is to the right). If v is negative, the particle is moving in the direction of negative r.

Relative to the sketch you drew in post #2, in which direction is the particle moving?
In which direction is the force of attraction?

so how do i know if its +r or -r, or +M or -M

how do i get negative part, its says it moves "under the influence of a mass M" does that mean that the direction is negative, because when it says "under the influence" i think it just means that its affected by it? not that its "under it"

i don't even know if the sketch is accurate. does it matter which way i sketch the r?

 

[Mark44]

so how do i know if its +r or -r, or +M or -M

The masses aren't going to be negative.

As for r, you need to be concerned with the direction the particle is moving. Its radial distance r is always going to be positive, but it's velocity can change sign, and that's what you need to consider. If r' > 0, the particle is moving away from O, in the direction of positive r. If r' < 0, the particle is moving toward O, in the direction of negative r. In your first post you have this:

"Initially P is at a distance of a from O when it is projected radially away from O at a speed u..."

how do i get negative part, its says it moves "under the influence of a mass M" does that mean that the direction is negative, because when it says "under the influence" i think it just means that its affected by it? not that its "under it"

The particle is projected away from O, but once it has reached a speed of u, the only force on it is the gravitational force due to mass M.

"Under the influence of" means that the particle is affected by the gravitation of mass M.

i don't even know if the sketch is accurate. does it matter which way i sketch the r?

Your sketch is accurate enough.

At time t = 0, in which direction is the particle moving? I.e., in the direction of positive r or negative r?
In which direction is the force of attraction? I.e., in the direction of positive r or negative r?

 

[cloud360]

The masses aren't going to be negative.

As for r, you need to be concerned with the direction the particle is moving. Its radial distance r is always going to be positive, but it's velocity can change sign, and that's what you need to consider. If r' > 0, the particle is moving away from O, in the direction of positive r. If r' < 0, the particle is moving toward O, in the direction of negative r. In your first post you have this:

"Initially P is at a distance of a from O when it is projected radially away from O at a speed u..."

The particle is projected away from O, but once it has reached a speed of u, the only force on it is the gravitational force due to mass M.

"Under the influence of" means that the particle is affected by the gravitation of mass M.
Your sketch is accurate enough.

At time t = 0, in which direction is the particle moving? I.e., in the direction of positive r or negative r?
In which direction is the force of attraction? I.e., in the direction of positive r or negative r?

when you say force of attraction you mean gravity.

so if the particle P moves away from O in positive direction, that means gravity is working against it in opposite direction.

Does this mean it is the gravity which is going to be negative as gravity awlays poitns downwards.


In conclusion, am i right to say in every single question like this, the gravity will be -G or -i in polar co ordinates (i.e pointing downwards)

so in this question should i just add a negative part and explain why i added it?is it that simple, or do i need to do some algebra?

 

[Mark44]

when you say force of attraction you mean gravity.

Yes, the only force in this problem is the force due to gravity.

so if the particle P moves away from O in positive direction, that means gravity is working against it in opposite direction.

Yes, and that's what I've been trying to get you to realize.

Does this mean it is the gravity which is going to be negative as gravity awlays poitns downwards.

No, I'm talking about the force that is due to gravity. Mass M exerts a force on particle P, but P also exerts a force on M. The force that particle P "feels" is in the direction of O, which is in the direction of negative r. P's velocity is initially in the direction of positive r.

Going back to your first post in this thread, what is force that P feels due to gravitational attraction? Which direction is that force pointing?

In conclusion, am i right to say in every single question like this, the gravity will be -G or -i in polar co ordinates (i.e pointing downwards)

We're not really dealing with polar coordinates here. Everything is occurring along a straight line. Don't think of gravity being negative or positive - focus on the force involved.

so in this question should i just add a negative part and explain why i added it?is it that simple, or do i need to do some algebra?

 

[cloud360]

Yes, the only force in this problem is the force due to gravity.
Yes, and that's what I've been trying to get you to realize.
No, I'm talking about the force that is due to gravity. Mass M exerts a force on particle P, but P also exerts a force on M. The force that particle P "feels" is in the direction of O, which is in the direction of negative r. P's velocity is initially in the direction of positive r.

Going back to your first post in this thread, what is force that P feels due to gravitational attraction? Which direction is that force pointing?
We're not really dealing with polar coordinates here. Everything is occurring along a straight line. Don't think of gravity being negative or positive - focus on the force involved.

Why is gravity not always -G in these questions?

also is it only -G, if it is acting against a particle moving in the positive r? would it be +G if the particle was moving with negative r?

 

[Mark44]

Why is gravity not always -G in these questions?

also is it only -G, if it is acting against a particle moving in the positive r? would it be +G if the particle was moving with negative r?

You're focussing on the wrong thing. G is just a constant (and it's positive). Focus on the force and which direction it points.

 

[cloud360]

You're focussing on the wrong thing. G is just a constant (and it's positive). Focus on the force and which direction it points.

so how can i justify adding a negative symbol. should i say

"the gravity is acting in the opposite direction of the particle"

so gravity = Gi, where i is a direction vector which is negative, so we have -G?

gravity would be positive, but -G just indicated the direction of the gravity?

I got my exam in 12 days i really need to know how i can add a negative symbol ASAP

 

[Mark44]

so how can i justify adding a negative symbol. should i say

"the gravity is acting in the opposite direction of the particle"

so gravity = Gi, where i is a direction vector which is negative, so we have -G?

gravity would be positive, but -G just indicated the direction of the gravity?

I got my exam in 12 days i really need to know how i can add a negative symbol ASAP

Forget the unit vectors, and forget -G. This problem is about one dimension.
You have F = ma = mr''

The gravitational pull from M causes P to accelerate in which direction - toward M or away from M?

 

[cloud360]

Forget the unit vectors, and forget -G. This problem is about one dimension.
You have F = ma = mr''

The gravitational pull from M causes P to accelerate in which direction - toward M or away from M?

Accelerates away from M, as it says in question? so what, what's the significance of this?

 

[Mark44]

No, it doesn't say this. What you posted says that the particle is projected away from M at a speed of u. After the particle is moving at a speed of u, the only force it feels is the gravitational attraction from mass M.

In what direction is the force acting on the particle?
In what direction is the particle's acceleration?

I'm hoping that you'll get this within the 12 days.

 

[cloud360]

I'm hoping that you'll get this within the 12 days.

i hoper so also, one of the last questions i posted took almost 14 days :)

In what direction is the force acting on the particle?
In what direction is the particle's acceleration?

I'm hoping that you'll get this within the 12 days.

The force is acting in the opposite direction of the vector r. the particle points towards the positive r. so the force points towards the negative r?

the particles acceleration is just in the direction of the vector r? or does the particle orbit the mass M or something.

Thanks again for your time, i really appreciate it

 

[Mark44]

No, it doesn't say this. What you posted says that the particle is projected away from M at a speed of u. After the particle is moving at a speed of u, the only force it feels is the gravitational attraction from mass M.

In what direction is the force acting on the particle?
In what direction is the particle's acceleration?

The force is acting in the opposite direction of the vector r. the particle points towards the positive r. so the force points towards the negative r?

Which direction does r point? What I'm looking for are answers like "toward M" or "away from M."

The particle doesn't point anywhere, but at the start, the particle is moving in the direction of positive r. IOW, away from M. And yes, the attraction force is in the direction of negative r; i.e., toward M.

the particles acceleration is just in the direction of the vector r? or does the particle orbit the mass M or something.

No and no. How can the particle's acceleration be in the opposite direction of the force due to gravity?

Also, this problem is strictly one-dimensional. It either moves in a straight line away from mass M, or it moves back in a straight line toward M. There is no orbiting going on.

 

[cloud360]

No, it doesn't say this. What you posted says that the particle is projected away from M at a speed of u. After the particle is moving at a speed of u, the only force it feels is the gravitational attraction from mass M.

In what direction is the force acting on the particle?
In what direction is the particle's acceleration?
Which direction does r point? What I'm looking for are answers like "toward M" or "away from M."

The particle doesn't point anywhere, but at the start, the particle is moving in the direction of positive r. IOW, away from M. And yes, the attraction force is in the direction of negative r; i.e., toward M.
No and no. How can the particle's acceleration be in the opposite direction of the force due to gravity?

Also, this problem is strictly one-dimensional. It either moves in a straight line away from mass M, or it moves back in a straight line toward M. There is no orbiting going on.

So, should i just add a negative symbol, saying gravity points towards M, so we are going backwards, so it points towards the negative M.

Or should i replace the work gravity above with force, if so, which force?

 

[Mark44]

How about answering my questions? I've asked more or less the same questions three times.

In what direction is the force acting on the particle?
In what direction is the particle's acceleration?

Or should i replace the work gravity above with force, if so, which force?

Yes, talk about the force.

Which force? As far as the particle is concerned, there is only one force.

 

[cloud360]

How about answering my questions? I've asked more or less the same questions three times.

In what direction is the force acting on the particle?
In what direction is the particle's acceleration?
Yes, talk about the force.

Which force? As far as the particle is concerned, there is only one force.

Force is acting in the opposite direction of the particle.

The particles acceleration is in the same direction as r

 

[Mark44]

Force is acting in the opposite direction of the particle.

Please be more precise. There are only two directions in this problem - toward M or away from M.

The particles acceleration is in the same direction as r

How can the direction of the particle's acceleration be different from the force applied to the particle? After all, the basic equation is F = ma.

If it seems that these posts go on and on - they do, for the reason that you have very many misconceptions about the physical concepts in this and the other problems you've posted. I keep asking questions in the hopes that you will begin understanding the ideas that are involved in this problem.

 

[cloud360]

Please be more precise. There are only two directions in this problem - toward M or away from M.

How can the direction of the particle's acceleration be different from the force applied to the particle? After all, the basic equation is F = ma.

If it seems that these posts go on and on - they do, for the reason that you have very many misconceptions about the physical concepts in this and the other problems you've posted. I keep asking questions in the hopes that you will begin understanding the ideas that are involved in this problem.

Ok the force moves towards M. but why, is it due to gravity. like how to the moon orbits the Earth due to gravity?

the direction of the particles acceleration is also towards m?

so r''=-r'' ?

 

[Mark44]

Ok the force moves towards M. but why, is it due to gravity. like how to the moon orbits the Earth due to gravity?

Yes, the force on the particle is toward mass M, and this force is due to gravitational attraction by M. Note the mass M also feels of force of the same magnitude but in the opposite direction. The force that M feels is due to gravitational attraction by particle P.

the direction of the particles acceleration is also towards m?

Of course. That follows directly from F = ma.

so r''=-r'' ?

How can the acceleration equal its own negative?

 

[cloud360]

Yes, the force on the particle is toward mass M, and this force is due to gravitational attraction by M. Note the mass M also feels of force of the same magnitude but in the opposite direction. The force that M feels is due to gravitational attraction by particle P.
Of course. That follows directly from F = ma.
How can the acceleration equal its own negative?

ok so if acceleration is towards m, then accelration is (r'')ȓ. where ȓ is a direction vector which is negative so we have a=-r'' ?

is this how we get a negative?

 

[Mark44]

We know that F = ma, and that F = mMG/r², with m being the mass of particle P and M being the mass of the other object.

Since the gravitational force on P is directed towards O, this force and P's acceleration are in the opposite direction of r, hence they are negative.

Therefore mr'' = - MG/r².

As I said before, this problem is strictly one-dimensional. In one direction quantities are positive; in the other direction they are negative.

 

[cloud360]

so was i right about a=(r'')ȓ

and that ȓ=negative so a=-r''

due to gravity acceleration points towards M, so ȓ is negative?

 

[Mark44]

so was i right about a=(r'')ȓ

No. This unit vector points in the same direction as r (away from M). The acceleration points the other way.

and that ȓ=negative so a=-r''

No, ȓ isn't negative.

due to gravity acceleration points towards M, so ȓ is negative?

Acceleration is toward M, yes, but ȓ is a unit vector in the other direction, away from M.

 

[cloud360]

No. This unit vector points in the same direction as r (away from M). The acceleration points the other way.
No, ȓ isn't negative.

Acceleration is toward M, yes, but ȓ is a unit vector in the other direction, away from M.

So would i be allowed to just make it negative saying its because gravity points towards M? so gravity has a negative direction vector?

 

[Mark44]

You can make the force negative because it is directed toward M.

Please stop talking about gravity and whether it's negative or positive. Instead, talk about the force that gravity exerts. Mass M exerts a force on P, so that from the perspective of P, the force is toward M. In the same manner, P exerts a force on M of equal magnitude, but in the opposite direction. From M's perspective, the force is toward P.

You persist in saying that gravity is working in this direction or that direction. In fact, gravity is working in both directions.

 

[cloud360]

You can make the force negative because it is directed toward M.

Please stop talking about gravity and whether it's negative or positive. Instead, talk about the force that gravity exerts. Mass M exerts a force on P, so that from the perspective of P, the force is toward M. In the same manner, P exerts a force on M of equal magnitude, but in the opposite direction. From M's perspective, the force is toward P.

You persist in saying that gravity is working in this direction or that direction. In fact, gravity is working in both directions.

Can you please kindly tell me if my answer to part b is correct or not

[PLAIN]http://img863.imageshack.us/img863/8066/2009b5bsol.gif

 

[Mark44]

You already asked me about this, and I answered in post #5 of this thread.
What you have in this post (#33) is mostly correct, but I would change a number of things that are still incorrect or unclear if I were turning this in.
1. "Particle of mass "m" moves from fixed point O --> r' = 0"
This is wrong. The problem statement says that P is initially at a distance of a from O and that it is moving at a speed of u.
2. "Anything distance = magnitude"
As I mentioned before, there is the concept of directed distance.
3. "Since there is displacement we have \int_{|r|}^r
The integral with limits but no integrand is meaningless. You should remove this whole line.
4. "|r'(0)| = r'(0) = a, as straight line"
This is wrong. |r'(0)| = u, the speed of the particle. |r(0)| = a.
What does "as straight line" mean?
5. "The 0 i.e. value inside bracket = gravity i.e. g/G when deriving Energy Equation"
I have no idea what you're trying to say here. Gravity is NOT a number and it is not 0.You should remove this whole line.
6. In the line where you integrate v dv you have an integral with limits of integration on the right side, but no limits of integration on the left side. You should simplify things by getting rid of the limits of integration on the right side.
Also, when you evaluate an integral that has integration limits, you don't get the arbitrary constant. Get rid of the limits of integration on the right side, but keep the constant.
7. There is no word "otherside." It should be "other side."
8. In the two lines where you go from 0.5v² to v² you have the same constant. When you multiply both sides of the equation by 2, the constant changes as well. If you look back at post #5 you'll see that I used a different constant, C' to indicate that we get a new constant.

 

[cloud360]

You already asked me about this, and I answered in post #5 of this thread.
What you have in this post (#33) is mostly correct, but I would change a number of things that are still incorrect or unclear if I were turning this in.
1. "Particle of mass "m" moves from fixed point O --> r' = 0"
This is wrong. The problem statement says that P is initially at a distance of a from O and that it is moving at a speed of u.
2. "Anything distance = magnitude"
As I mentioned before, there is the concept of directed distance.
3. "Since there is displacement we have \int_{|r|}^r
The integral with limits but no integrand is meaningless. You should remove this whole line.
4. "|r'(0)| = r'(0) = a, as straight line"
This is wrong. |r'(0)| = u, the speed of the particle. |r(0)| = a.
What does "as straight line" mean?
5. "The 0 i.e. value inside bracket = gravity i.e. g/G when deriving Energy Equation"
I have no idea what you're trying to say here. Gravity is NOT a number and it is not 0.You should remove this whole line.
6. In the line where you integrate v dv you have an integral with limits of integration on the right side, but no limits of integration on the left side. You should simplify things by getting rid of the limits of integration on the right side.
Also, when you evaluate an integral that has integration limits, you don't get the arbitrary constant. Get rid of the limits of integration on the right side, but keep the constant.
7. There is no word "otherside." It should be "other side."
8. In the two lines where you go from 0.5v² to v² you have the same constant. When you multiply both sides of the equation by 2, the constant changes as well. If you look back at post #5 you'll see that I used a different constant, C' to indicate that we get a new constant.

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

for |r’(0)|=u , what is this , is it r? or gravity i.e should i sub r=0 or G=0?
[PLAIN]http://img231.imageshack.us/img231/1016/solf.gif

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

 

[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&#039;
where C' = 2C

At t = 0, v(0) = u and r(0) = a, so
u² = 2MG/a + C'
\Rightarrow C&#039; = 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.