# Physics help-Newtons law of gravitation

Watch
Announcements
#1
I don't understand the link between these equations and the difference between them - can somebody help pls?
0
2 years ago
#2
Do you not understand it in physics or mathematical terms? 0
#3
(Original post by dreamerz11)
Do you not understand it in physics or mathematical terms? physics basically self teaching physics atm-just did my A level maths this year
0
2 years ago
#4
I don't know if this is what you're asking for or if I'm too late, but:

- The first equation is the gravitational force between two masses (e.g. between two planets). The reason the equation is this is due to the definition of Newton's law of gravitation that states that 2 point masses attract each other with a force proportional to the product of their masses but inversely proportional to the square of their separation. The gravitational constant (6.67x10^-11) is the constant of proportionality in this case.

- The second equation is the value of the gravitational field strength in a radial field, meaning the value of gravity in one planet's atmosphere (the Earth for instance). I guess if I had to explain it to myself I'd say it's just a rearrangement of the F=ma formula, substituting the value of acceleration to be the gravitational field strength (acceleration due to gravity). The reason why it is Gm(1) / r^2 is because F/m(2) would equal Gm(1)m(2) / m(2)r^2 so the m(2) would cancel out.

I hope this helps in any way (I also hope this is right lol)
0
#5
(Original post by Kane987)
I don't know if this is what you're asking for or if I'm too late, but:

- The first equation is the gravitational force between two masses (e.g. between two planets). The reason the equation is this is due to the definition of Newton's law of gravitation that states that 2 point masses attract each other with a force proportional to the product of their masses but inversely proportional to the square of their separation. The gravitational constant (6.67x10^-11) is the constant of proportionality in this case.

- The second equation is the value of the gravitational field strength in a radial field, meaning the value of gravity in one planet's atmosphere (the Earth for instance). I guess if I had to explain it to myself I'd say it's just a rearrangement of the F=ma formula, substituting the value of acceleration to be the gravitational field strength (acceleration due to gravity). The reason why it is Gm(1) / r^2 is because F/m(2) would equal Gm(1)m(2) / m(2)r^2 so the m(2) would cancel out.

I hope this helps in any way (I also hope this is right lol)
Ah thanks a lot!!!!!! This makes a lot of sense You explain it so well!!
Could you pls kindly help with a related problem to do this?
and got confused as to why at 7:21, M does not change but at 9:30 M does change?
1
#6
(Original post by Kane987)
I don't know if this is what you're asking for or if I'm too late, but:

- The first equation is the gravitational force between two masses (e.g. between two planets). The reason the equation is this is due to the definition of Newton's law of gravitation that states that 2 point masses attract each other with a force proportional to the product of their masses but inversely proportional to the square of their separation. The gravitational constant (6.67x10^-11) is the constant of proportionality in this case.

- The second equation is the value of the gravitational field strength in a radial field, meaning the value of gravity in one planet's atmosphere (the Earth for instance). I guess if I had to explain it to myself I'd say it's just a rearrangement of the F=ma formula, substituting the value of acceleration to be the gravitational field strength (acceleration due to gravity). The reason why it is Gm(1) / r^2 is because F/m(2) would equal Gm(1)m(2) / m(2)r^2 so the m(2) would cancel out.

I hope this helps in any way (I also hope this is right lol)
Could you pls kindly help with a related problem to do this if you get the chance?
and got confused as to why at 7:21, M does not change but at 9:30 M does change?
Im self teaching physics lol
0
2 years ago
#7
I'm sorry, I'm not quite sure what you mean So I'll have a go at just answering the questions myself XD

- So at 7:21, they're asking for the value ofthe gravitational force acting on 2 different bodies.

A). Okay, first thing we have to do is write down the equation and all the quantities we have, since we're using the equation F = GMm/r^2 (M = m(1) and m = m(2))
So:
G = 6.67x10^-11
M = 5.97x10^24 (Mass of Earth)
m = 450000 (Mass of ISS - the SI unit of mass is kilograms which is why I didn't convert it to grams)

r = the radius of the Earth + the distance away from Earth the object is, so it becomes (6.37x10^6 + 400x10^3)

So you do F = (6.67x10^-11)(5.97x10^24)(450000) divided by (6.37x10^6 + 400x10^3)^2

which equals 3.9x10^6 on my calculator (essentially what the answer is in the video).

B). For this part, you use all the same quantities, but you substitute the value of m for the mass of the astronaut which gives me 564N on my calculator.

I hope that made sense For the next question at 9:30,

I think the term they've used there 'grav.accel' means gravitational field strength hence they used the equation g = GM / R^2 (R is the radius of the planet)

Okay, the reason M has changed is because it has told us it is 1/5 the mass of Earth (so originally M would be the mass of Earth, but now it is the mass of Earth x 0.2)

So from the question:
G = 6.67x10^-11
M = 1/5M or 0.2M
R = 1/2R or 0.5R

So substituting the values in:

g = 0.2GM/(0.5R)^2

(0.5R)^2 = 0.25R^2, subbing this in you get: g = 0.2GM/0.25R^2

you can move the numbers in this fraction to get g = 0.2/0.25 x GM/R^2
(0.2/0.25 = 0.8) so g = 0.8 x GM/R^2

Now, since we are comparing to the Earth where the g = 9.81, this means that GM/R^2 = 9.81, so you can sub it into the equation to get:

g on this random planet = 0.8 x 9.81 = 7.848

Sorry for taking so long and that this is a bit of a mess, I'm new to this site and need to learn to format better lol 1
#8
(Original post by Kane987)
I'm sorry, I'm not quite sure what you mean So I'll have a go at just answering the questions myself XD

- So at 7:21, they're asking for the value ofthe gravitational force acting on 2 different bodies.

A). Okay, first thing we have to do is write down the equation and all the quantities we have, since we're using the equation F = GMm/r^2 (M = m(1) and m = m(2))
So:
G = 6.67x10^-11
M = 5.97x10^24 (Mass of Earth)
m = 450000 (Mass of ISS - the SI unit of mass is kilograms which is why I didn't convert it to grams)

r = the radius of the Earth + the distance away from Earth the object is, so it becomes (6.37x10^6 + 400x10^3)

So you do F = (6.67x10^-11)(5.97x10^24)(450000) divided by (6.37x10^6 + 400x10^3)^2

which equals 3.9x10^6 on my calculator (essentially what the answer is in the video).

B). For this part, you use all the same quantities, but you substitute the value of m for the mass of the astronaut which gives me 564N on my calculator.

I hope that made sense For the next question at 9:30,

I think the term they've used there 'grav.accel' means gravitational field strength hence they used the equation g = GM / R^2 (R is the radius of the planet)

Okay, the reason M has changed is because it has told us it is 1/5 the mass of Earth (so originally M would be the mass of Earth, but now it is the mass of Earth x 0.2)

So from the question:
G = 6.67x10^-11
M = 1/5M or 0.2M
R = 1/2R or 0.5R

So substituting the values in:

g = 0.2GM/(0.5R)^2

(0.5R)^2 = 0.25R^2, subbing this in you get: g = 0.2GM/0.25R^2

you can move the numbers in this fraction to get g = 0.2/0.25 x GM/R^2
(0.2/0.25 = 0.8) so g = 0.8 x GM/R^2

Now, since we are comparing to the Earth where the g = 9.81, this means that GM/R^2 = 9.81, so you can sub it into the equation to get:

g on this random planet = 0.8 x 9.81 = 7.848

Sorry for taking so long and that this is a bit of a mess, I'm new to this site and need to learn to format better lol Ah thanks a lot!! I really really appreciate your time!!!!
For the first problem, where did you get M = 5.97x10^24 (Mass of Earth) -it doesn't state this in the problem

Also, really sorry I didn't explain myself properly - so At 7:18 the lady says that G doesn't change,Mass of the Earth doesn't change and the only thing that changes is the distance to the centre of the Earth.

However, at 9:16 the lady says that G doesn't change, the Mass changes and the radius changes.

what is bothering me here is that in both questions it's to do with finding something between two objects so I am finding it slightly hard to understand the reasoning for the above stated.

I am really sorry if I am repeating anything you have previously kindly explained.

Could you please kindly let me know your thoughts if you d=get the chance - no worries if not possible 0
2 years ago
#9
No problem! Okay,

- For the first problem, I used the mass of Earth to make it easier for me (that is what the mass of Earth actually is) as during my A-Levels a few months ago it had the value in the equation booklet we got in the exam meaning we were allowed to use it. However, the question would probably ask you to use the information like that given to you so I can try to explain it with only the information in the question if you need me to (I was just cutting corners because I prefer to do it that way XD) - As for the difference between 7:21 and 9:28,

So, in both cases, G won't change. G won't change in any circumstance as it is just a number like 3 or 28.2, it's just referred to as the gravitational constant and given the symbol G. Think of it like pi, it has the symbol for it, but it really just means the number 3.14....... and so on XD

As for why M changes, at 7:21 it just talks about the Earth so you put M as the mass of the Earth. At 9:28 she says M changes because in the question it says the mass of the planet is 1 fifth the mass of Earth. What she means by M changing is just that if it had just said Earth, M would have been the mass of Earth. Since it now says 1 fifth the mass of Earth, M has changed to the mass of Earth x 1/5.
I think it may have just been the wording that may have confused for a second. M has not 'changed' in the general sense, she means it has changed relative to the previous question because in the question it says the mass is different. Is that clearing it up at all? (I realise my explanation is long haha)

As for the distance to the centre of the Earth. Again, she doesn't mean the radius of Earth has changed, she means that the radius you're working with has changed relative to if you were using the radius of Earth. What I mean is, if the question said the radius was the same as Earth, you would say R = the radius of the Earth. Since the question says the Radius is 1/2 of Earth's, the radius you're working with has 'changed' so that r = 1/2 the radius of the Earth.

Does that help at all? Oh hang on! I almost pressed submit lol, another thing I forgot to mention, both questions aren't asking about 2 objects. The questions A and B at 7:21 are because they are asking for the gravitational force between A) the ISS and the Earth and B) the astronaut and the Earth. However, the question at 9:30 is only asking about the gravitational field strength of one planet (aka what value of gravity you would experience if you were on the planet), it is just giving you the values you need to calculate this as a fraction of what they are on Earth to make it harder 0
#10
(Original post by Kane987)
No problem! Okay,

- For the first problem, I used the mass of Earth to make it easier for me (that is what the mass of Earth actually is) as during my A-Levels it had the value in the equation booklet we got in the exam meaning we were allowed to use it. However, the question would probably ask you to use the information like that given to you so I can try to explain it with only the information in the question if you need me to (I was just cutting corners because I prefer to do it that way XD) - As for the difference between 7:21 and 9:28,

So, in both cases, G won't change. G won't change in any circumstance as it is just a number like 3 or 28.2, it's just referred to as the gravitational constant and given the symbol G. Think of it like pi, it has the symbol for it, but it really just means the number 3.14....... and so on XD

As for why M changes, at 7:21 it just talks about the Earth so you put M as the mass of the Earth. At 9:28 she says M changes because in the question it says the mass of the planet is 1 fifth the mass of Earth. What she means by M changing is just that if it had just said Earth, M would have been the mass of Earth. Since it now says 1 fifth the mass of Earth, M has changed to the mass of Earth x 1/5.
I think it may have just been the wording that may have confused for a second. M has not 'changed' in the general sense, she means it has changed relative to the previous question because in the question it says the mass is different. Is that clearing it up at all? (I realise my explanation is long haha)

As for the distance to the centre of the Earth. Again, she doesn't mean the radius of Earth has changed, she means that the radius you're working with has changed relative to if you were using the radius of Earth. What I mean is, if the question said the radius was the same as Earth, you would say R = the radius of the Earth. Since the question says the Radius is 1/2 of Earth's, the radius you're working with has 'changed' so that r = 1/2 the radius of the Earth.

Does that help at all? Oh hang on! I almost pressed submit lol, another thing I forgot to mention, both questions aren't asking about 2 objects. The questions A and B at 7:21 are because they are asking for the gravitational force between A) the ISS and the Earth and B) the astronaut and the Earth. However, the question at 9:30 is only asking about the gravitational field strength of one planet (aka what value of gravity you would experience if you were on the planet), it is just giving you the values you need to calculate this as a fraction of what they are on Earth to make it harder Ah yess that's it!! I get it all - I cannot thank you enough for your help!! I have a question about something else but I am not sure how to word it lol so I am going to revise over it before I ask you haha
Once again, thank you very much!
Have a very goodnight! .. err morning haha
0
2 years ago
#11
Brilliant! I'm not the best at physics so I was actually really happy I could help at all (I'm waiting for my results this year and I keep saying to my family I'll be happy with anything over a U because I didn't take the initiative like you and use YouTube and places like here to ask for help, oh well!)

Anyway, I can certainly try to help with any other question (and I will apologise in advance if I can't because I just haaate certain topics - especially the harder parts of electricity lol XD). Oh! and it is almost 02:00 so it may take longer to respond from a mobile instead of the desktop and if I go to sleep before you ask the question I'll get back to you as soon as possible (but I know deep down I'm gonna fight sleep for as long as possible lol XD)

Anyway, glad to help! 0
#12
(Original post by Kane987)
Brilliant! I'm not the best at physics so I was actually really happy I could help at all (I'm waiting for my results this year and I keep saying to my family I'll be happy with anything over a U because I didn't take the initiative like you and use YouTube and places like here to ask for help, oh well!)

Anyway, I can certainly try to help with any other question (and I will apologise in advance if I can't because I just haaate certain topics - especially the harder parts of electricity lol XD). Oh! and it is almost 02:00 so it may take longer to respond from a mobile instead of the desktop and if I go to sleep before you ask the question I'll get back to you as soon as possible (but I know deep down I'm gonna fight sleep for as long as possible lol XD)

Anyway, glad to help! Ah you're the best! And no way defo at least an A grade - I am sure you have aced it!I am just forcing myself to revise lol And pls sleep - I am so sorry if you are awake bc of my ******ed questions lol
I swear I am so grateful for your help!! 0
2 years ago
#13
Ahaha, I was gonna be awake anyway browsing so I thought I'd see if there was anything productive to do, so I stumbled upon this question You should've seen how many questions I missed out and did the wrong thing on my exam (I literally could not think of one right thing I did on paper 3 and I'm refusing to look at the unofficial mark scheme lol, so I'll be happy if it is anything over a U as it means it wasn't 2 years for nothing XD) so an A is a pipe dream lol!

Anyway, I will take your advice and sleep so I don't fall out of bed at 12 lol XD
0
#14
(Original post by Kane987)
Ahaha, I was gonna be awake anyway browsing so I thought I'd see if there was anything productive to do, so I stumbled upon this question You should've seen how many questions I missed out and did the wrong thing on my exam (I literally could not think of one right thing I did on paper 3 and I'm refusing to look at the unofficial mark scheme lol, so I'll be happy if it is anything over a U as it means it wasn't 2 years for nothing XD) so an A is a pipe dream lol!

Anyway, I will take your advice and sleep so I don't fall out of bed at 12 lol XD
Ah crap - oh well, fingers crossed you've done better than you expected!
And yeh lol Goodnight 0
#15
(Original post by Kane987)
Ahaha, I was gonna be awake anyway browsing so I thought I'd see if there was anything productive to do, so I stumbled upon this question You should've seen how many questions I missed out and did the wrong thing on my exam (I literally could not think of one right thing I did on paper 3 and I'm refusing to look at the unofficial mark scheme lol, so I'll be happy if it is anything over a U as it means it wasn't 2 years for nothing XD) so an A is a pipe dream lol!

Anyway, I will take your advice and sleep so I don't fall out of bed at 12 lol XD
Hey sorry to bother again So at around 3:36 of this vid and onwards I am a bit confused https://www.youtube.com/watch?v=ks1B1_umFk8

So the formula is confusing me the most .. so how do you know what value for the charge you put in ?

Could you pls kindly help with this? Nw if not possible 0
2 years ago
#16
Hmm, I'm not so great with electricity, but I'll try my best (keeping in mind I'm just hoping what I'm saying is right lol)

I'm not so great with explaining the video, so I'll explain it in my own way if that's alright So!

For starters, I'll be referring to q(1) as Q....and q(2) as q

Absolute Electric Potential is the work done per unit charge in bringing a small positive test charge (q) from infinity to a point in the field.
We have 2 equations for this which are:

V = ΔW/q or V = ΔE/q

So we can rearrange the second equation to get E = qV where E is the electric potential energy, q is the charge at the point in the field away from big Q and V is the electric potential.

So at 3:36 in the video, he tells us that the big Qis in the middle of the screen, and that the little q is near the far left bottom. He says that the electric potential is 100JC-1 and that the charge of little q is 2C....he wants us to find the electrical potential energy.

So we use the equation E = qV and sub in our values to get E = 2 x 100 = 200J of energy.

So at 5:00 in the video, he tells us another equation for electric potential which is V = K x Q/r
Notice how this equation uses a big Q instead of a little q? That's because this equation only concerns the charge that is causing the electric potential. In the video, before he placed that charge there the electric potential is 0, but after he puts it there, it causes electric potential to be created, so that's why we're using that one (it's just what we're told in the equation).

Now, just like the gravitational constant we talked about yesterday 'G', 'K' also never changes as it is just a number (which he says is 9x109 in the video).

So why is this the equation we use? (Just incase you may ask). Well, it comes from breaking up the equation for electric potential.
So, my definition at the top says:

'the work done........in bringing a small positive test charge from infinity to a point' (I've omitted the 'per unit charge' for now)

The equation we learned for the work done in bringing it from infinity to a point is:

W = K x Qq/r

But since the definition says 'per unit charge' to get the value of electric potential, we can say that:

V = K x Qq/rqwhich would equal V = K x Q/r

Where K is the constant, Q is the charge causing the electric potential, q is the charge placed at another point and r is the distance between the two charges.

So at 10:00 in the video, he starts talking about a change in electric potential or electric potential difference, the symbol being ΔV and the unit being volts (electric potential difference).

So the simple equation to calculate this if you had both electric potentials is ΔV = V2 - V1 (V2 being the second point and V1 being the first point)
There is another equation to use but it links a little differently so I'll only explain that part of electric potential if needs be (the part where you work out the work done in bringing the charge to a point between 2 charges, but that'll just complicate what I've told you here)

Has this helped at all or am I just rambling lol Edit: I tried to attach pictures of the equations to make it easier but they are only showing up at the bottom of my post, just ignore them haha
0
#17
(Original post by Kane987)
Hmm, I'm not so great with electricity, but I'll try my best (keeping in mind I'm just hoping what I'm saying is right lol)

I'm not so great with explaining the video, so I'll explain it in my own way if that's alright So!

For starters, I'll be referring to q(1) as Q....and q(2) as q

Absolute Electric Potential is the work done per unit charge in bringing a small positive test charge (q) from infinity to a point in the field.
We have 2 equations for this which are:

V = ΔW/q or V = ΔE/q

So we can rearrange the second equation to get E = qV where E is the electric potential energy, q is the charge at the point in the field away from big Q and V is the electric potential.

So at 3:36 in the video, he tells us that the big Qis in the middle of the screen, and that the little q is near the far left bottom. He says that the electric potential is 100JC-1 and that the charge of little q is 2C....he wants us to find the electrical potential energy.

So we use the equation E = qV and sub in our values to get E = 2 x 100 = 200J of energy.

So at 5:00 in the video, he tells us another equation for electric potential which is V = K x Q/r
Notice how this equation uses a big Q instead of a little q? That's because this equation only concerns the charge that is causing the electric potential. In the video, before he placed that charge there the electric potential is 0, but after he puts it there, it causes electric potential to be created, so that's why we're using that one (it's just what we're told in the equation).

Now, just like the gravitational constant we talked about yesterday 'G', 'K' also never changes as it is just a number (which he says is 9x109 in the video).

So why is this the equation we use? (Just incase you may ask). Well, it comes from breaking up the equation for electric potential.
So, my definition at the top says:

'the work done........in bringing a small positive test charge from infinity to a point' (I've omitted the 'per unit charge' for now)

The equation we learned for the work done in bringing it from infinity to a point is:

W = K x Qq/r

But since the definition says 'per unit charge' to get the value of electric potential, we can say that:

V = K x Qq/rqwhich would equal V = K x Q/r

Where K is the constant, Q is the charge causing the electric potential, q is the charge placed at another point and r is the distance between the two charges.

So at 10:00 in the video, he starts talking about a change in electric potential or electric potential difference, the symbol being ΔV and the unit being volts (electric potential difference).

So the simple equation to calculate this if you had both electric potentials is ΔV = V2 - V1 (V2 being the second point and V1 being the first point)
There is another equation to use but it links a little differently so I'll only explain that part of electric potential if needs be (the part where you work out the work done in bringing the charge to a point between 2 charges, but that'll just complicate what I've told you here)

Has this helped at all or am I just rambling lol Edit: I tried to attach pictures of the equations to make it easier but they are only showing up at the bottom of my post, just ignore them haha
Ah I don't even know how to thank you! I really really appreciate your help!!!
"So at 5:00 in the video, he tells us another equation for electric potential which is V = K x Q/r
Notice how this equation uses a big Q instead of a little q? That's because this equation only concerns the charge that is causing the electric potential. In the video, before he placed that charge there the electric potential is 0, but after he puts it there, it causes electric potential to be created, so that's why we're using that one (it's just what we're told in the equation). "
So referring to this in the video at 2:43 it says before we put it there the V value was 100 but you said the electrical potential was 0 so kinda confused about this.
Could you pls kindly lemme know whenever you can  0
2 years ago
#18
Ahh, I think I forgot to mention what time.

I meant that the electric potential was 0 at the start of the video before drawing anything in. When he draws in the big Q with the green circle around it (the first thing he draws) electric potential is created.

I think what you were referring to was the little q in the bottom left (blue dot) but I meant the green one in the middle 0
#19
(Original post by Kane987)
Ahh, I think I forgot to mention what time.

I meant that the electric potential was 0 at the start of the video before drawing anything in. When he draws in the big Q with the green circle around it (the first thing he draws) electric potential is created.

I think what you were referring to was the little q in the bottom left (blue dot) but I meant the green one in the middle Thank you so much! But you know the little q, doesn't it also produce an electric potential which can like affect the Big Q.

Oh wait isn't it an electric field and not an electric potential ..electrical potential enenrgy
crap im getting it all mixed up lol
0
2 years ago
#20
Anyone wants to do A levels Physics and Maths group study ? We can help each other
0
X

new posts Back
to top
Latest
My Feed

### Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

see more

### See more of what you like onThe Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

### Poll

Join the discussion

#### Current uni students - are you thinking of dropping out of university?

Yes, I'm seriously considering dropping out (49)
16.23%
I'm not sure (8)
2.65%
No, I'm going to stick it out for now (103)
34.11%
I have already dropped out (4)
1.32%
I'm not a current university student (138)
45.7%