# Thread: HELP - I am getting Totall Lost - Gravity from a Single Object.

1. Riiiight - I keep getting confusing and different Answers.

Gravity from a Single Object.

Taking the Formula g = G x M / r ^2.

However - I have Masses in Kg and Distances in Km.

So - do I take g = 1/10^6 x G x M / r^2.

I am TRYING to get a Factor for the Gravity of the Sun, at the Orbit of the Planet Mercury.

So - Sun Mass M = 1,412 x 10^18

Mercury Orbit = 4.6 x 10^ Km.

This gives me the Value of 3.76 x 10^9 x G.

This seems VERY high to me.

Is this Right - Or am I Very Stupid ??  2.

3. Yes, you should use meters to avoid getting confused (you should also obtain the correct value of G in this system).

Sun mass is, anyway, approximately 2*10^30 kg.
What you've written about the orbit is the perihelion (for what I see in Wikipedia). The mean distance (think it is a mean) is 5.8*10^7 km
This gives a value of g = 5.95*10^8 G, if I'm not wrong, which is equal to g = 0.04 m/s^2

The one of the Earth, is 0.006 m/s^2, and it's a order of magnitude less because it is farther than Mercury, why does it seem too much to you?  4. Originally Posted by fred91 Yes, you should use meters to avoid getting confused (you should also obtain the correct value of G in this system).

Sun mass is, anyway, approximately 2*10^30 kg.
What you've written about the orbit is the perihelion (for what I see in Wikipedia). The mean distance (think it is a mean) is 5.8*10^7 km
This gives a value of g = 5.95*10^8 G, if I'm not wrong, which is equal to g = 0.04 m/s^2

The one of the Earth, is 0.006 m/s^2, and it's a order of magnitude less because it is farther than Mercury, why does it seem too much to you?
Thanks for the Help.

Damn Big Values - I will just have to live with it.
I know - it LOOKS Big - because the Value is G is so Tiny - G = 6.67384 x 10^-11.

My problem - but this is the Third Time I have had to change the Calculations - and thus the Rules, in the last Two Years.

__________________________________________________ _____________________________

I also had to make a Powers of 10 Correction to my Spreadsheet.

The Mass given for the Sun / r^2 ( where r = 5.791 x 10^7 / 2 )
Using Sun Mass = 1.9881 x 10^30
These are Units of Kg and Km.

NOTE - 1.9891 x 10^30 is the value for Sun Mass in Kg - from Wiki.

So - Agreed - I need to Correct for Km to K.
As the Radius, in Km is "on the bottom" and Squared - the Value for the Result needs to be Corrected by 1 / 1 x 10^6.

HOWEVER - This gives a Value of 5.931 x 10^9 x G.

To get your given Answer of "near to" 5.95 x 10^8 - I have to make it 10 times SMALLER.

This gives me a Value of 5.931 x 10^9 x G.  5. Originally Posted by Karakris  Originally Posted by fred91 Yes, you should use meters to avoid getting confused (you should also obtain the correct value of G in this system).

Sun mass is, anyway, approximately 2*10^30 kg.
What you've written about the orbit is the perihelion (for what I see in Wikipedia). The mean distance (think it is a mean) is 5.8*10^7 km
This gives a value of g = 5.95*10^8 G, if I'm not wrong, which is equal to g = 0.04 m/s^2

The one of the Earth, is 0.006 m/s^2, and it's a order of magnitude less because it is farther than Mercury, why does it seem too much to you?
Thanks for the Help.

Damn Big Values - I will just have to live with it.
I know - it LOOKS Big - because the Value is G is so Tiny - G = 6.67384 x 10^-11.

My problem - but this is the Third Time I have had to change the Calculations - and thus the Rules, in the last Two Years.

__________________________________________________ _____________________________

I also had to make a Powers of 10 Correction to my Spreadsheet.

The Mass given for the Sun / r^2 ( where r = 5.791 x 10^7 / 2 )
Using Sun Mass = 1.9881 x 10^30
These are Units of Kg and Km.

NOTE - 1.9891 x 10^30 is the value for Sun Mass in Kg - from Wiki.

So - Agreed - I need to Correct for Km to K.
As the Radius, in Km is "on the bottom" and Squared - the Value for the Result needs to be Corrected by 1 / 1 x 10^6.

HOWEVER - This gives a Value of 5.931 x 10^9 x G.

To get your given Answer of "near to" 5.95 x 10^8 - I have to make it 10 times SMALLER.

This gives me a Value of 5.931 x 10^9 x G.

The first thing to do is make sure that all your values are in meters, kilograms, and sec before you ever put them into the equation. you are much less likely to make a mistake this way than trying to do the conversion after plugging into the equation.

Secondly, why did you use r = 5.791 x 10^7 / 2 ?

The mean distance of Mercury from the Sun is 5.791e7 km or 5.791e10 meters, this is the value you should be using for r.

Thus you should get

(6.6738e-11)(1.9881e30)/ (5.791e10)^2 = 0.03956 m/s^2 for the acceleration due to gravity at Mercury's orbit.

You can independently check this by the following:

Mercury has an orbital period of 87.97 days or 7,600,608 sec long. The circumference of a circular orbit of radius 5.791e10 m is 3.639e11 m. This gives an orbital speed of 3.638e11/7,600,608 = 47864m/s

The centripetal acceleration for an object circling at this radius and speed is found by v^2/r = 47864^2/5.791e10 = 0.03956 m/s^2  6. Thanks Janus - but putting in a 1/10^6 factor works just as well for me.

Space Disances in Metres is just silly, for me - as all our Calculations are in Kg and Km.

I am not NOW halving the Value for Radius of 5.97 x 10^7.

I need to know this Value in terms of values of the Gravitational Constant G ( 6.67384 x 10^-11 ).

For this Calculation - I get a Value at roughly average Mercury Orbit of 6.262 x 10^9 x G.

That is gonna be used as my Limiting Value -
Inertialess Fields ( Krallin Fields ) can only work when the External g Force ( in this case, Gravity of Sun ) is equal to or less than 6.5 x 10^9 x G.
This wikll be our Workable Limit - for using Inertialless Fields for ANY Purpose - but in particlar for Space Flight to the Planet, Mercury.

Sorry - YES - it is for a Science Fiction Game - but I do need to have accurate REAL Figures.
That is why I put it in the REAL Physics Section.

Thanks to Everyone, for all of your Help.  7. As the Formula is -

g = G x M / r^2.

If I am using Km instead of metres on the "bottom" - then the Correction fo these different Units is that the Answer will be
1,000 ^2 = 1 x 10^6 Too Big.

So - I Corrected the Answer by a Factor of 1 / 10^6.
This seems perfectly reasonable to me - as I already have the Orbital Info. in Km in the Spreadsheet.

I am not typing these Figures in by hand - they are one of the Calculations from a Spreadsheet - which uses "reasonable" Values for Astronomical Calculations.
Giving Orbits of Planets in METRES is Not "reasonable" - that is why Wiki does not give them in Metres, but in Km.

Also - Mercury is just ONE of the Planets - my Spreadsheet is providing values for ALL of the Planets, including Trans-Neptunians and Kuiper Belt Objects.

I have also used these Calculations, in the Spreadsheet - with r = Radius of Planet, to Calculate the Surface Gravity, where it is not given.
My Calculations must be fairly correct - because I DO get 1.981 m / s^2 for Earth.  8. Originally Posted by Karakris Thanks Janus - but putting in a 1/10^6 factor works just as well for me.

Space Disances in Metres is just silly, for me - as all our Calculations are in Kg and Km.

I am not NOW halving the Value for Radius of 5.97 x 10^7.

I need to know this Value in terms of values of the Gravitational Constant G ( 6.67384 x 10^-11 ).

For this Calculation - I get a Value at roughly average Mercury Orbit of 6.262 x 10^9 x G.

That is gonna be used as my Limiting Value -
Inertialess Fields ( Krallin Fields ) can only work when the External g Force ( in this case, Gravity of Sun ) is equal to or less than 6.5 x 10^9 x G.
This wikll be our Workable Limit - for using Inertialless Fields for ANY Purpose - but in particlar for Space Flight to the Planet, Mercury.

Sorry - YES - it is for a Science Fiction Game - but I do need to have accurate REAL Figures.
That is why I put it in the REAL Physics Section.

Thanks to Everyone, for all of your Help.
It doesn't really make any sense to use g-force as your limiting factor, since anything in free fall ( which includes orbits) are at 0g no matter how close to the Sun they get. If you want to base the limit on gravity, it would make more sense to base it on gravitational differential or tidal forces. I.E. the gravitational differential can't be greater than a certain value over the extent of the field or that the gradient can't be steeper than a given value. In the first case, the size of the field would be limited by proximity to a gravity source ( the closer you are, the smaller the field can be). In the second case, the field existence is limited by the value.  9. Originally Posted by Janus  Originally Posted by Karakris Thanks Janus - but putting in a 1/10^6 factor works just as well for me.

Space Distances in Metres is just silly, for me - as all our Calculations are in Kg and Km.

I am not NOW halving the Value for Radius of 5.97 x 10^7.

I need to know this Value in terms of values of the Gravitational Constant G ( 6.67384 x 10^-11 ).

For this Calculation - I get a Value at roughly average Mercury Orbit of 6.262 x 10^9 x G.

That is gonna be used as my Limiting Value -
Inertialess Fields ( Krallin Fields ) can only work when the External g Force ( in this case, Gravity of Sun ) is equal to or less than 6.5 x 10^9 x G.
This will be our Workable Limit - for using Inertialless Fields for ANY Purpose - but in particular for Space Flight, for instance to the Planet Mercury.

Sorry - YES - it is for a Science Fiction Game - but I do need to have accurate REAL Figures.
That is why I put it in the REAL Physics Section.

Thanks to Everyone, for all of your Help.
It doesn't really make any sense to use g-force as your limiting factor, since anything in free fall ( which includes orbits) are at 0g no matter how close to the Sun they get. If you want to base the limit on gravity, it would make more sense to base it on gravitational differential or tidal forces. I.E. the gravitational differential can't be greater than a certain value over the extent of the field or that the gradient can't be steeper than a given value. In the first case, the size of the field would be limited by proximity to a gravity source ( the closer you are, the smaller the field can be). In the second case, the field existence is limited by the value.
My thoughts and the Principles are -
That even if it an Object is in Orbit around the Sun - it will still FEEL the Force of g from the Sun - otherwise it would go off in a Tangent.
Acording to Newton - an object will continue in straight line motion, Unless it is acted on by an External Force.
This g Force from the Sun makes the Orbit of Mercury depart from a straight line, always falling inwards towards the Sun -
But the g Force must BE THERE, and it must have an effect, or Mercury would fly off in a tangent.

Our Inertialless Fields DO require Power in proportion to the Mass of the Object, and / or in proportion to the Volume of space which they enclose - whichever is greater. I have to work out the Power Ratios, shortly.

They can make everything INSIDE the Field at Zero Inertia ( only whilst they are in operation ), and free from External Gravity. However - they cannot effect things OUTSIDE of the Field.
When the External g Force becomes greater than a certain value - the Inertialess Field just Fails.
When an Inertialess Field is turned off - the original or Intrinsic Inertia is restored, it has only been Suspended, not destroyed.

We have Limits on the operation of our Fast FTL Drives, which operate inside an Inertialess Field.
This Limit is once again related to the external g Force - for Sol, this means that they will NOT operate at all, when as near to the Sun as Mars Orbit.
It is the Sun's Gravity which limits this.
We obviously cannot use our FTL Drives when on any Planet, Moon or Asteroid - unless it was a very tiny piece of rock.  Bookmarks
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