Can someone explain why this happens -Gpotential physics topic help needed!!! Watch

Batman2k1
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If there are 2 planets, lets say Earth and mars,
now field lines are radial meaning it gets weaker, as you go further from earth or mars
now why at the centre of the distances between earths and mars (to their radius) is the gravatational potential = 0???
must be something i am not getting,
gp is gpe per unit mass?
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homemadeclock
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it's not, because the gravitational potential is not a vector. But work either has to be put in or is done by the system when an object is moved from a point, therefore, it has a sign. For an object placed at point P work would have to be done against the pull of both masses to move it to infinity - their result is, therefore, the sum of the potentials and the potential will be an expression of the potentials, not zero
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Eimmanuel
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(Original post by Batman2k1)
If there are 2 planets, lets say Earth and mars,
now field lines are radial meaning it gets weaker, as you go further from earth or mars
now why at the centre of the distances between earths and mars (to their radius) is the gravatational potential = 0???
must be something i am not getting,
gp is gpe per unit mass?
I think you confused gravitational field strength g with gravitational potential V. But they are related:
 g = - \dfrac{dV }{dr}

Note that your example of a system (Earth and Mars), the gravitational field strength is NOT zero at the midpoint because the mass of Earth (5.97 × 1024 kg) is about 10 times more than that of the Mars (6.39 × 1023 kg).

The gravitational field strength of the Earth (m1) and Mars (m2) system versus the separation would look like the picture as shown below. The zero gravitational field strength position would be nearer to Mars (m2).
Name:  Gravitational_potential_field 2.JPG
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Eimmanuel
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(Original post by Batman2k1)
If there are 2 planets, lets say Earth and mars,
now field lines are radial meaning it gets weaker, as you go further from earth or mars
now why at the centre of the distances between earths and mars (to their radius) is the gravatational potential = 0???
must be something i am not getting,
gp is gpe per unit mass?

I cannot find the gravitational potential versus the separation of Earth (m1) and Mars (m2) system graph, so I use the gravitational potential versus the separation of Earth and Moon system graph to illustrate the relationship between gravitational field strength g and gravitational potential V.

The gravitational potential versus the separation of the Earth and Moon system graph is shown below. If you take the negative gradient of the graph of gravitational potential, you can produce the gravitational field strength graph which looks similar to the graph in post #3.
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Batman2k1
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If my understanding is correct, let's say i am in the centre distance measured between mars radius to earths radius. So if i was placed at the centre, whats happening is, mars is pulling me back, as well as earth too, to move a point to infinity. So work is done, and it is negative as it is against the gravatational forces. As there is equal work done which equates to GP (GPE per unit mass) hence 0 GP - is this right?
(Original post by homemadeclock)
it's not, because the gravitational potential is not a vector. But work either has to be put in or is done by the system when an object is moved from a point, therefore, it has a sign. For an object placed at point P work would have to be done against the pull of both masses to move it to infinity - their result is, therefore, the sum of the potentials and the potential will be an expression of the potentials, not zero
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Batman2k1
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Ok, let's see if i understood this. g=f/m just refers to a mass of an object. g=GM/r^2 refers to the magnitude of the gfs in a radial field. This corresponds to earth having a higher gfs due to a larger mass of course.

Therefore, if i am the mass at the point closer to m2 that is mars. The GFS =0 as the Gforce pulls me strongly from Earth than Mars, so obviously, to counteract that closer to Mars. The why is probably in the GP bit.
if GP = GPE per unit mass at a point. So more work is done from earth's gravitational pull than mars. So if i am at the position closer to m2,
then the work done of earth would equate to the work done by mar's gravitational pull, because GFS gets weaker when you increase separation distance of course.

the graph demonstrated from moon and earths GPE against separation - i find confusing.
If you take the negative gradient of the graph of gravitational potential, you can produce the gravitational field strength graph which looks similar to the graph in post #3 - not seeing the picture - may need 2 go back 2 the basics.


(Original post by Eimmanuel)
I cannot find the gravitational potential versus the separation of Earth (m1) and Mars (m2) system graph, so I use the gravitational potential versus the separation of Earth and Moon system graph to illustrate the relationship between gravitational field strength g and gravitational potential V.

The gravitational potential versus the separation of the Earth and Moon system graph is shown below. If you take the negative gradient of the graph of gravitational potential, you can produce the gravitational field strength graph which looks similar to the graph in post #3.
Name:  Gravitational_potential 3.JPG
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Eimmanuel
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(Original post by Batman2k1)
Ok, let's see if i understood this. g=f/m just refers to a mass of an object. g=GM/r^2 refers to the magnitude of the gfs in a radial field. This corresponds to earth having a higher gfs due to a larger mass of course.

Therefore, if i am the mass at the point closer to m2 that is mars. The GFS =0 as the Gforce pulls me strongly from Earth than Mars, so obviously, to counteract that closer to Mars. The why is probably in the GP bit.
if GP = GPE per unit mass at a point. So more work is done from earth's gravitational pull than mars. So if i am at the position closer to m2,
then the work done of earth would equate to the work done by mar's gravitational pull, because GFS gets weaker when you increase separation distance of course.

the graph demonstrated from moon and earths GPE against separation - i find confusing.
If you take the negative gradient of the graph of gravitational potential, you can produce the gravitational field strength graph which looks similar to the graph in post #3 - not seeing the picture - may need 2 go back 2 the basics.
First of all, I cannot really say that I understand what are you writing. If I "really understand" your writing correctly, it seems that you have issue understanding the fundamental definition of gravitational potential and its implication(s). I would try to expand when I have the time.

There is nothing wrong in finding the graph confusing but you need to say what is confusing.
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homemadeclock
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I agree, i am trying to understand what you are saying too
(Original post by Eimmanuel)
First of all, I cannot really say that I understand what are you writing. If I "really understand" your writing correctly, it seems that you have issue understanding the fundamental definition of gravitational potential and its implication(s). I would try to expand when I have the time.

There is nothing wrong in finding the graph confusing but you need to say what is confusing.
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Batman2k1
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confuseception lol!
Eimmanuel Yeah alright, i guess i would need a lesson gpotential implication and definition again. I don't know how it works. Strip it down to the basics. I am unsure about the graph why does vmoon start at 60 separation. shouldn't it be at 0m as increase the value up to 0?

(Original post by homemadeclock)
I agree, i am trying to understand what you are saying too
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Batman2k1
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can you clarify why potential = 0?
(Original post by homemadeclock)
it's not, because the gravitational potential is not a vector. But work either has to be put in or is done by the system when an object is moved from a point, therefore, it has a sign. For an object placed at point P work would have to be done against the pull of both masses to move it to infinity - their result is, therefore, the sum of the potentials and the potential will be an expression of the potentials, not zero
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Eimmanuel
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(Original post by Batman2k1)
confuseception lol!
Eimmanuel Yeah alright, i guess i would need a lesson gpotential implication and definition again. I don't know how it works. Strip it down to the basics. I am unsure about the graph why does vmoon start at 60 separation. shouldn't it be at 0m as increase the value up to 0?

I am astonished by your question “why does vmoon start at 60 separation. shouldn't it be at 0m as increase the value up to 0?”

How can Moon and Earth occupy the same point in space? They can occupy the same point in space, provided collision takes place.


A typical gravitational potential curve due to a point mass would look like an electric potential curve due to a point charge which is negatively charged.
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Batman2k1
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Oh so the earth position starts at 0m initially? Ok so r represents the separation distance between the moon and earth. Initially, they are separate. No that cant be right.

So at 0m, they are not separated at all, space point in space. So if the GP of the earth seems to be a very high negative number at 0m, then that would mean it takes a lot of work done against the gravitational forces because if i was at point of mass on earth, it takes a lot of energy to go against whats pulling me down? But the moon at 0m, doesn't take that much GP because it's grav forces are way weaker as it has a smaller mass, so its being pulled by the Earth, therefore if i was on the moon, doesn't that much energy, to go against the moons gravity at that point?

Where does the earth position initially start? It's really confusing - can u help me out
Does that make sense? Easy to depict and understand? Any fatal mistake here?

(Original post by Eimmanuel)
I am astonished by your question “why does vmoon start at 60 separation. shouldn't it be at 0m as increase the value up to 0?”

How can Moon and Earth occupy the same point in space? They can occupy the same point in space, provided collision takes place.


A typical gravitational potential curve due to a point mass would look like an electric potential curve due to a point charge which is negatively charged.
Name:  Electric_potential_negative.gif
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Eimmanuel
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(Original post by Batman2k1)
Oh so the earth position starts at 0m initially? Ok so r represents the separation distance between the moon and earth. Initially, they are separate. No that cant be right.

So at 0m, they are not separated at all, space point in space. So if the GP of the earth seems to be a very high negative number at 0m, then that would mean it takes a lot of work done against the gravitational forces because if i was at point of mass on earth, it takes a lot of energy to go against whats pulling me down? But the moon at 0m, doesn't take that much GP because it's grav forces are way weaker as it has a smaller mass, so its being pulled by the Earth, therefore if i was on the moon, doesn't that much energy, to go against the moons gravity at that point?

Where does the earth position initially start? It's really confusing - can u help me out
Does that make sense? Easy to depict and understand? Any fatal mistake here?
If you really want someone to help you, I would suggest that you put in the effort to construct your sentence for other to understand your problem better.
I am not sure now whether you are trying to understand gravitational potential energy (GPE) or gravitational potential (GP): GPE and GP are two different physical quantities but they are related.

Based on your writing, you seem to be describing gravitational potential energy and then at some other places, you turn to gravitational potential. I cannot really make sense of what are you trying to tell me.
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Eimmanuel
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(Original post by Batman2k1)
Ok, let's see if i understood this. g=f/m just refers to a mass of an object. g=GM/r^2 refers to the magnitude of the gfs in a radial field. This corresponds to earth having a higher gfs due to a larger mass of course.

Therefore, if i am the mass at the point closer to m2 that is mars. The GFS =0 as the Gforce pulls me strongly from Earth than Mars, so obviously, to counteract that closer to Mars. The why is probably in the GP bit.
if GP = GPE per unit mass at a point. So more work is done from earth's gravitational pull than mars. So if i am at the position closer to m2,
then the work done of earth would equate to the work done by mar's gravitational pull, because GFS gets weaker when you increase separation distance of course.

the graph demonstrated from moon and earths GPE against separation - i find confusing.
If you take the negative gradient of the graph of gravitational potential, you can produce the gravitational field strength graph which looks similar to the graph in post #3 - not seeing the picture - may need 2 go back 2 the basics.

I highly suggested that you watched the following videos first. Again it is ok not to understand by the first time of watching them.

Gravitational & Electric Fields - A-level & GCSE Physics
https://www.youtube.com/watch?v=KfZIjNY7Cp0

Gravitational Potential A-Level Physics
https://www.youtube.com/watch?v=4u1dcqxuDcI


Before going to talking about gravitational potential, I would talk about gravitational potential energy (GPE) first.

I would not write all the things that can be found online, so I would refer you to have a look at the following link. You don't need to understand the maths (the calculus part) but you need to know the results and implications.
https://phys.libretexts.org/Bookshel...d_Total_Energy

There are a few points that I would like you pay attention to.


 U = - \dfrac{GMm}{r} ---- (1)

The formula tells us that gravitational potential energy is a property of the system of two particles rather than of either particle alone. There is no way to divide this energy and say that so much belongs to one particle and so much to the other. However, if Mm, as is true for Earth (mass M) and a ball (mass m), a lot of physics textbooks would have the saying of “the gravitational potential energy of the ball”, so is it wrong?

Not that the textbook is wrong but you need to be careful and know why it is ok to write in such way. When the ball moves near the surface of the Earth, the changes in the gravitational potential energy of the ball–Earth system appear almost entirely as changes in the kinetic energy of the ball, since changes in the kinetic energy of Earth are too small to be measured.

When we speak of the gravitational potential energy of bodies of comparable mass, however, we have to be careful to treat them as a system.
Name:  GPE_Curve.png
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When you are given a GPE curve as shown above, you need to know the GPE value is meant for the system of two masses NOT one mass.
There is also a reference value: when the 2 masses are very very very far apart, the gpe of the system is zero.
When you are given a GPE value say at r = r1, you need to interpret the value with respect to the reference point (r = ∞).
In physics, it is the change in GPE that has meaning.
The "r" in equation (1) is the separation between 2 masses.

------------
Next, if we define gravitational potential (GP) at a point is the work done per unit mass in bringing a mass from infinity to the point (typical A level physics definition). From equation (1), we can write gravitational potential of a source mass M as


 V = - \dfrac{GM}{r} ---- (2)

So what is gravitational potential (GP)? Before saying about what is GP, let see how GP is applied. I know you may think I have defined GP, the answer is yes and no. Why not just continue reading the rest of the writing.

Considers 2 points (r1 and r2) in space measured from the source mass M and the gravitational potential at these 2 points are V1 and V2 respectively. The gravitational potential difference between these 2 points is related to the work done W12 by the gravitational force by source mass M on a mass m when the mass moves from 1 to 2.


 \Delta V = V_2 - V_1 = - \dfrac{W_{12}}{m} ---- (3)

Using (3) and conservation of energy allows us to determine whether a particle of mass m speeds up or slows down as it moves through that region.
Say that the gravitational potential at r2 is more negative than that of r1 (from (2), we would know that r1 is further away from the source mass) and we place a particle of mass m at r1, work done by the gravitational force exerted on m by the source mass M is positive in moving the particle from r2 to r1. The particle of mass m speeds up as it moves towards lower gravitational potential: gravitational potential energy of the system (source mass M and mass m) is transformed into kinetic energy of the mass m assuming that Mm.

I would stop here for today. I need to think of how to "transform eqn (3)" without much maths.
The following video has the essence of what I would like to reach.
https://www.youtube.com/watch?v=xrQCPYsoBKk
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Eimmanuel
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Continued from post # 14:

Where would I like to go from here? I like to show the connection between GP and gravitational field.

Suppose we move a test mass m0 from point 1 to point 2 in one dimension along a gravitational field line in a gravitational field g.

The work done by the gravitational force F = m0g on the test mass is

 W_{12} = \int_{1}^{2}m_0 \vec{\bold{g}} \cdot  d\vec{\bold{r}}

and using (3), we have


 \Delta V = V_2 - V_1 = - \dfrac{W_{12}}{m_0} =-\dfrac{\int_{1}^{2}\vec{\bold{F}  }\cdot d\vec{\bold{r}}}{m_0}=-\dfrac{\int_{1}^{2}m_0\vec{\bold  {g}}\cdot d\vec{\bold{r}}}{m_0}=-\int_{1}^{2}\vec{\bold{g}}\cdot d\vec{\bold{r}}

OR


 \Delta V = V_2 - V_1 = -\int_{1}^{2}\vec{\bold{g}}\cdot d\vec{\bold{r}} ---- (4)

(For simplicity, you can think of an integral as an area under a curve: area under the g versus r curve between 1 and 2)

Similar to GPE, we may wish to find the gravitational potential at a point, relative to some chosen reference potential, rather than the potential difference given by Eqn (4). If we choose the reference point to be at infinity and define V = 0 as the reference (similar to the reference point in GPE), then Eqn (4) gives for the potential at point P:

 V_P = -\int_{\infty}^{P}\vec{\bold{g}} \cdot d\vec{\bold{r}} ---(5)

Instead of writing Eqn (5) using gravitational field, we can write the work done by the gravitational field (W) on a test mass m0 as that test mass moves in from infinity to point P and define the gravitational potential at any point in the gravitational field to be

 V = - \dfrac{W_{\infty}}{m_0} ---- (5a)

This means that the gravitational potential is a property of the source mass and characteristic of the gravitational field due to the source mass, independent of a particle of mass m that may be placed in the field. The gravitational potential and gravitational field are not two distinct entities but, instead, two different perspectives or two different mathematical representations of how source mass(es) alter the space around them. (You may need to read these 2 sentence a few times or comes back to these after study gravitational field in detail.)

Last but not least, let us apply eqn (3) to find the gravitational potential due to a system of two mass.

If we have a system of Earth and Moon which is separated by about 60rE and we want to find the gravitational potential due to the Earth and Moon at a distance r = 20rE (no special reason for choosing this point). We just add the gravitational potential due to the Earth and the Moon,

 V = - \dfrac{GM_E}{20R_E} + (-\dfrac{GM_m}{60R_E - 20R_E})

I would refer to the graph shown in post #4 now. What this implies is that to find the “resultant” gravitational potential curve VT (blue curve) due to the system of Earth and Moon, we can just superimpose (or “add”) the gravitational potential curve due to the Earth VEarth (red curve) and gravitational potential curve VMoon (green curve) due to the Moon.

Mathematically, this means

 V_T(r) = - \dfrac{GM_E}{r} - \dfrac{GM_m}{60R_E - r}

where r is the point between the Earth and Moon and is measured with respect to the centre of the Earth.

I do not claim that this is an easy way of understanding gravitational potential. The mathematical equations between GPE and GP may look similar but the interpretation of them is not the same. So be careful and alert of the pitfall. If you have problems in understanding any of the writing, please post your questions with proper sentence construction.
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Batman2k1
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Thank you! I will come back to you after i have read the sources,

and try to be aware of my sentence construction as best as i can.
I am curious, is it the organisation and layout of points i make whilst discussing X physics topic?
(Original post by Eimmanuel)
Continued from post # 14:

Where would I like to go from here? I like to show the connection between GP and gravitational field.

Suppose we move a test mass m0 from point 1 to point 2 in one dimension along a gravitational field line in a gravitational field g.

The work done by the gravitational force F = m0g on the test mass is

 W_{12} = \int_{1}^{2}m_0 \vec{\bold{g}} \cdot  d\vec{\bold{r}}

and using (3), we have


 \Delta V = V_2 - V_1 = - \dfrac{W_{12}}{m_0} =-\dfrac{\int_{1}^{2}\vec{\bold{F}  }\cdot d\vec{\bold{r}}}{m_0}=-\dfrac{\int_{1}^{2}m_0\vec{\bold  {g}}\cdot d\vec{\bold{r}}}{m_0}=-\int_{1}^{2}\vec{\bold{g}}\cdot d\vec{\bold{r}}

OR


 \Delta V = V_2 - V_1 = -\int_{1}^{2}\vec{\bold{g}}\cdot d\vec{\bold{r}} ---- (4)

(For simplicity, you can think of an integral as an area under a curve: area under the g versus r curve between 1 and 2)

Similar to GPE, we may wish to find the gravitational potential at a point, relative to some chosen reference potential, rather than the potential difference given by Eqn (4). If we choose the reference point to be at infinity and define V = 0 as the reference (similar to the reference point in GPE), then Eqn (4) gives for the potential at point P:

 V_P = -\int_{\infty}^{P}\vec{\bold{g}} \cdot d\vec{\bold{r}} ---(5)

Instead of writing Eqn (5) using gravitational field, we can write the work done by the gravitational field (W) on a test mass m0 as that test mass moves in from infinity to point P and define the gravitational potential at any point in the gravitational field to be

 V = - \dfrac{W_{\infty}}{m_0} ---- (5a)

This means that the gravitational potential is a property of the source mass and characteristic of the gravitational field due to the source mass, independent of a particle of mass m that may be placed in the field. The gravitational potential and gravitational field are not two distinct entities but, instead, two different perspectives or two different mathematical representations of how source mass(es) alter the space around them. (You may need to read these 2 sentence a few times or comes back to these after study gravitational field in detail.)

Last but not least, let us apply eqn (3) to find the gravitational potential due to a system of two mass.

If we have a system of Earth and Moon which is separated by about 60rE and we want to find the gravitational potential due to the Earth and Moon at a distance r = 20rE (no special reason for choosing this point). We just add the gravitational potential due to the Earth and the Moon,

 V = - \dfrac{GM_E}{20R_E} + (-\dfrac{GM_m}{60R_E - 20R_E})

I would refer to the graph shown in post #4 now. What this implies is that to find the “resultant” gravitational potential curve VT (blue curve) due to the system of Earth and Moon, we can just superimpose (or “add”) the gravitational potential curve due to the Earth VEarth (red curve) and gravitational potential curve VMoon (green curve) due to the Moon.

Mathematically, this means

 V_T(r) = - \dfrac{GM_E}{r} - \dfrac{GM_m}{60R_E - r}

where r is the point between the Earth and Moon and is measured with respect to the centre of the Earth.

I do not claim that this is an easy way of understanding gravitational potential. The mathematical equations between GPE and GP may look similar but the interpretation of them is not the same. So be careful and alert of the pitfall. If you have problems in understanding any of the writing, please post your questions with proper sentence construction.
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