Problem with Newton's law of universal gravitation

[Harald Linke]

Hello everyone,

my name is Harald Tamer Linke and I'm 16 years old. At the moment I'm trying to get a little deeper into astrophysics and for doing such I'm trying to learn some formulas f.e.: the equation for universal gravitation: 

F = G*(m1*m2/r^2)

But my problem is that when I try using this equation for an example, instead of Newton I get Newton/m^2 as result. Here is my equation:

The gravitation between two objects (70kg and 80kg) with a distance of 3 meters:

F = G * (m1 * m2/r^2) = (9.81N/kg) * 150kg * (80kg * 70kg/(3m)^2)

= (9.81N/kg) * 150kg * (5600kg/9m^2)

= (9.81N/kg) * 150kg * 622,22..kg/m^2

= (9.81N/kg) * 93333,3...kg/m^2 

= 915.600 N/m^2

 

Can someone maybe take the time and explain my mistake(s) to me? 

 

Thank you in advance.

 

 

Edited January 13, 2019 by Harald Linke

[Ghideon]

Are you using correct constant? G vs g?

"g" should not be confused with "G", which is the standard symbol for the gravitational constant.

 

(short answer, I’m using phone)

 

Edited January 13, 2019 by Ghideon

[studiot]

No problem this is simple to resolve.

 

:)

 

There are two gravitational constants.

The one which appears in Newton's Law of Gravity, which you started with, is given the symbol 'big G' (as you did) and has a value

6.674×10⁻¹¹ N. kg. m²

This is called the Universal Gravitational Constant.

Note the units.

https://en.wikipedia.org/wiki/Gravitational_constant

 

The second one is called the acceleration due to gravity and is independent of mass.
It is given the symbol 'ittle g'

This is the one you have used the value of in your working.

You would nomally use this in kinematic equations such as

v = u + at with the acceleration = 9.81 metres per second²

Does this help?

[Harald Linke]

1 minute ago, Ghideon said:

Are you using correct constant? G vs g?

"g" should not be confused with "G", which is the standard symbol for the gravitational constant.

 

(short answer, I’m using phone)

 

damn... the gravitational constant is correct but I somehow (don't ask me why) rep

 

1 minute ago, studiot said:

No problem this is simple to resolve.

 

 

There are two gravitational constants.

The one which appears in Newton's Law of Gravity, which you started with, is given the symbol 'big G' (as you did) and has a value

6.674×10⁻¹¹ N. kg. m²

This is called the Universal Gravitational Constant.

Note the units.

https://en.wikipedia.org/wiki/Gravitational_constant

 

The second one is called the acceleration due to gravity and is independent of mass.
It is given the symbol 'ittle g'

This is the one you have used the value of in your working.

You would nomally use this in kinematic equations such as

v = u + at with the acceleration = 9.81 metres per second²

Does this help?

yeah... that totaly helps! to be honest I knew that I had to use the gravitational constant but somehow (bad research) got the equation G = g*m and used that. thanks for your fast answer!

[studiot]

Note when using these equations that Newton's formula involving big G is only part of the story.

It implies that that two bodies cannot be in mechanical equilibrium (Think about it)

So there must be other forces acting in any real universe system to allow stable solar systems, orbits and so on.

 

The acceleration due to gravity starts with the assumption that the mass of the Earth is so large compared to a rocket or whatever that all the activity can be attributed to the rocket and the Earth stays still.

 

[J.C.MacSwell]

12 hours ago, studiot said:

Note when using these equations that Newton's formula involving big G is only part of the story.

It implies that that two bodies cannot be in mechanical equilibrium (Think about it)

So there must be other forces acting in any real universe system to allow stable solar systems, orbits and so on.

 

The acceleration due to gravity starts with the assumption that the mass of the Earth is so large compared to a rocket or whatever that all the activity can be attributed to the rocket and the Earth stays still.

 

Why do you feel that assumption is needed? Any acceleration would be with respect to an inertial frame...not the Earth. The Earth is just a convenient approximation much of the time.

[studiot]

43 minutes ago, J.C.MacSwell said:

Why do you feel that assumption is needed? Any acceleration would be with respect to an inertial frame...not the Earth. The Earth is just a convenient approximation much of the time.

Read Harald's personal introduction.

So how would you explain it at that level?

Would inertial or any  frames be on your syllabus?

 

 

[J.C.MacSwell]

7 minutes ago, studiot said:

Read Harald's personal introduction.

So how would you explain it at that level?

Would inertial or any  frames be on your syllabus?

 

 

Inertial...with an explanation as to what can be considered a convenient approximation, and why.

I'm not sure Harald doesn't already have a working understanding of that...just got a little mixed up on the G.

[mathematic]

Where did the 150kg in your original equation come from?

[studiot]

10 hours ago, mathematic said:

Where did the 150kg in your original equation come from?

That is a very good question

+1

11 hours ago, J.C.MacSwell said:

Inertial...with an explanation as to what can be considered a convenient approximation, and why.

I'm not sure Harald doesn't already have a working understanding of that...just got a little mixed up on the G.

 

Inertial frames and inertia are not on the A level syllabus (or similar), which Harald has probably just started.

Here is all a standard text says about Newton's law of gravitation and big G v little g.
Note it simple says Newton proposed the inverse square law (which is true he did working from Kepler's observed data).
This is substantially less than used to be on the syllabus about this.

However I think part of the confusion arises from the modern habit/mania of renaming everything to something unintuitive.

So the acceleration due to gravity, g has become The gravitational field strength.
I have highlighted the important implications at the top of the third page.
Note the apparantly different units.

 

Edited January 14, 2019 by studiot

[Raider5678]

1 hour ago, studiot said:

However I think part of the confusion arises from the modern habit/mania of renaming everything to something unintuitive.

1

They got tired of being mocked for using uncreative names by non-scientists, so the scientists got their revenge by naming them creative but completely unintuitive things.

On 1/13/2019 at 6:00 AM, Harald Linke said:

F = G * (m1 * m2/r^2) = (9.81N/kg) * 150kg * (80kg * 70kg/(3m)^2)

 

This equation seems to have an error.

Where did the * 150kg * come from?

(m1 * m2/r^2) = * (80kg * 70kg/(3m)^2)

Which then you simply need to put in the gravitational constant again at the beginning. 

But seemingly, 150kg has appeared from nowhere, which is slightly confusing and may be contributing to your error.

This equation should be:

F = G * (m1 * m2/r^2) = 6.674×10⁻¹¹ N. kg. m² * (80kg * 70kg/(3m)^2)

6.674×10⁻¹¹ N. kg. m² * (5600kg/9m^2)

6.674×10⁻¹¹ N. kg. m² * 622.22...kg/m^2

[J.C.MacSwell]

1 hour ago, studiot said:

 

 

Inertial frames and inertia are not on the A level syllabus (or similar), which Harald has probably just started.

 

I would suggest that should be taught much earlier if that is the case. (I am not familiar with your levels, but he is 16)

[taeto]

1 hour ago, Raider5678 said:

But seemingly, 150kg has appeared from nowhere, which is slightly confusing and may be contributing to your error.

Absolutely not. Clearly 150kg is the combined mass of the two objects .

[swansont]

16 hours ago, J.C.MacSwell said:

Why do you feel that assumption is needed? Any acceleration would be with respect to an inertial frame...not the Earth. The Earth is just a convenient approximation much of the time.

The earth being still makes it an inertial frame, and also (typically) the frame of the observer

[Raider5678]

1 hour ago, taeto said:

Absolutely not. Clearly 150kg is the combined mass of the two objects .

I understand that, but where did combined mass enter into the equation?

F = G*(m1*m2/r^2)

That's the basic equation.

The equation that was presented was:

F = G*(m1+m2)*(m1*m2/r^2)

ANd that equations answer is going to be significantly different.

[J.C.MacSwell]

1 hour ago, swansont said:

The earth being still makes it an inertial frame, and also (typically) the frame of the observer

...and because it is not still, you need pseudo forces, like coriolis, to compensate, as it spins and wobbles around the Sun.

But as I said, it can be a reasonable approximation much of the time.

[studiot]

On 1/13/2019 at 11:00 AM, Harald Linke said:

But my problem is that when I try using this equation for an example, instead of Newton I get Newton/m^2 as result.

 

25 minutes ago, Raider5678 said:

I understand that, but where did combined mass enter into the equation?

Of course it does.

Harald simply made a mistake. (though please don't ask me why).

However he seems to have understood the idea of dimensional analysis since he says he got newtons per square metre instead of newtons for his force, and realised something was up the swannee.

This much is good.

Remember that acceleration is rate of change of velocity ie rate of change of metres per second or metres per second per second or M⁰L¹T^-2

and force is defined by Newtons second law as mass x acceleration which has dimensions M¹ x M⁰L¹T^-2 = MLT^-2

3 hours ago, J.C.MacSwell said:

I would suggest that should be taught much earlier if that is the case. (I am not familiar with your levels, but he is 16)

 

I agree, but sadly that is no longer the case.

My guess is that Harald hails from Norway so I don't know their system by I haven't heard it is in advance of the UK's.

The bottom line is that too much class time is devoted to astrophysics and cosmology, which are currently more sexy, and not enough to the basics, IMHO.

[Janus]

On 1/14/2019 at 9:18 AM, Raider5678 said:

I understand that, but where did combined mass enter into the equation?

F = G*(m1*m2/r^2)

That's the basic equation.

Harald explained this in an earlier thread:  He mistakenly thought that G was the equivalent of g*m.  He also though that m in this equation was the combined mass.  So he substituted

9.81 N/kg * (70kg+80kg) for G in the gravitational force equation. (9.81N/kg = 9.81m/s2 )

Edited January 15, 2019 by Janus

[Raider5678]

6 hours ago, Janus said:

Harald explained this in an earlier thread:  He mistakenly thought that G was the equivalent of g*m.  He also though that m in this equation was the combined mass.  So he substituted

9.81 N/kg * (70kg+80kg) for G in the gravitational force equation. (9.81N/kg = 9.81m/s2 )

Ah, I didn't catch that.