How to calculate gravitational constant

[robertroman10]

I don't seem to understand the m3 km and seconds used in the gravitational constant. If one body weighed 100kg and another 75kg could someone show me step by step how to solve this? I am 13 so please try to explain it at this level. Thanks.

 

[tiny-tim]

hi robertroman10!

(you mean kg, not km )

I don't seem to understand the m3 km and seconds used in the gravitational constant.

F = Gm₁m₂/r²,

so G has dimensions of force times distance-squared over mass squared

since force has dimensions of mass times acceleration, ie mass times length over time squared, that all multiplies out to distance-cubed over mass times time-squared

If one body weighed 100kg and another 75kg could someone show me step by step how to solve this? I am 13 so please try to explain it at this level.

the force between them, measured in Newtons, would be G times 100 times 75 divided by the distance (in metres) squared

see http://en.wikipedia.org/wiki/Gravitational_constant" for some more details

 

[robertroman10]

Okay so I was looking on the wikipedia like and it showed this as an alternate equation

G=6.674 * 10^-11 N(m/kg)2 since this one doesn't have time it seems simpler but where do they get the m and kg from? Is the meters the distance between them? and how do I calculate the Newtons (im infering N means Newtons)

 

[DrGreg]

Okay so I was looking on the wikipedia like and it showed this as an alternate equation

G=6.674 * 10^-11 N(m/kg)2 since this one doesn't have time it seems simpler but where do they get the m and kg from? Is the meters the distance between them? and how do I calculate the Newtons (im infering N means Newtons)

tiny-tim has already given you the answer:

$$F = G \frac{m_1 m_2}{r^2}$$​

F is the force (in Newtons)
m₁ and m₂ are the two masses (in kilograms)
r is the distance between them (in metres)
G is 6.67428 × 10⁻¹¹ (in N m² / kg²)

 

[robertroman10]

yes but I am asking how do we solve for N and kg?

 

[jhae2.718]

yes but I am asking how do we solve for N and kg?

I'm not quite sure what you mean by this. Gravitational force, as stated already, is given by
$$F_G = \frac{Gm_1m_2}{r^2}$$.

The gravitational constant G has units N m²/kg²*. This is just a number, and is what is called a constant of proportionality for gravitational force. What that means is that we know the force of gravity between two objects is given by the product of the masses divided by the square of the distance between them. However, we need to multiply by a constant--G--to get the actual magnitude of the force.

Now, to solve for force, if you know the masses and distance between them, you can just plug it into the equation. To find a mass if you know one mass, distance, and force, you can rearrange the equation:
$$m_1 = \frac{F_Gr^2}{Gm_2}$$

*Notice that the m²/kg² in the units for G will cancel out, leaving Newtons, which is the unit of force, and is the correct unit for the answer of the equation.

 

[robertroman10]

So if the gravitational constant is a constant what does it equal?

 

[jtbell]

Try Google.

 

[robertroman10]

I already did. What I want to know is if the gravitational constant is a specific number (for ex. 34.5) or does it vary based on the mass and distance of two bodies. If it varies what I want to know is how to solve the equation please show me step by step how you would solve it (if the constant varies and is not one particular number)

 

[tiny-tim]

it's a specific number, like c or h or the charge of a electron

 

[robertroman10]

So is 6.67428 * 10^-11 m3 kg^-1 S^-2 that number? and if it is, then how would you work this out let's say...on a calculator? because as far as my knowledge goes, you can't type in kg, m, and s's on a calculator

 

[HallsofIvy]

The value of G was measured by Henry Cavendish in 1797- 98:
http://en.wikipedia.org/wiki/Cavendish_experiment

 

[tiny-tim]

… how would you work this out let's say...on a calculator? because as far as my knowledge goes, you can't type in kg, m, and s's on a calculator

the whole point of using a consistent set of units, like the SI system, is that you don't have to bother about the units

 

[robertroman10]

Okay so on your first answer/post (something like "It would be G times 100 times 75 divided by the distance squared" what number would you replace G with? What is the 'final' answer? Thanks again

 

[FtlIsAwesome]

I would like to note that the Newton is actually a combination of units. It has seconds in it, and therefore includes time.

http://en.wikipedia.org/wiki/Newton_(unit )

The Newton is the SI unit for force; it is equal to the amount of net force required to accelerate a mass of one kilogram at a rate of one meter per second squared.

 

[jhae2.718]

Okay so on your first answer/post (something like "It would be G times 100 times 75 divided by the distance squared" what number would you replace G with? What is the 'final' answer? Thanks again

If you're using MKS, G = 6.67e-11 N m^2/kg^2

 

[FtlIsAwesome]

So is 6.67428 * 10^-11 m3 kg^-1 S^-2 that number? and if it is, then how would you work this out let's say...on a calculator? because as far as my knowledge goes, you can't type in kg, m, and s's on a calculator

Okay so on your first answer/post (something like "It would be G times 100 times 75 divided by the distance squared" what number would you replace G with? What is the 'final' answer? Thanks again

I believe he is asking how to use the numbers in a calculator without the units, and still get the correct answer.

 

[DrGreg]

Okay so on your first answer/post (something like "It would be G times 100 times 75 divided by the distance squared" what number would you replace G with? What is the 'final' answer? Thanks again

If the masses are both measured in kilograms, and the distance is measured in metres, and you want the force to be in Newtons, then G is 6.67428 × 10⁻¹¹. Always.

 

[robertroman10]

okay now I am lost... can someone just please show how to solve this STEP BY STEP thanks for your help.

 

[jtbell]

Start with the equation

$$F = G \frac {m_1 m_2} {r^2}$$

Substitute the value of G that Dr. Greg and others have given. Substitute the two masses, in kilograms. Substitute the distance, in meters. When you calculate the result, you get the force in Newtons.

Calculators don't deal with units. You have to keep them straight yourself. The value of G that we're using here assumes that force is measured in Newtons, distance is in meters, and mass is in kilograms. If you wanted to use (for example) force in pounds, distance in furlongs, and mass in slugs, then you'd have to use a different value of G.

 

[tiny-tim]

G has dimensions of … distance-cubed over mass times time-squared

The value of G that we're using here assumes that force is measured in Newtons, distance is in meters, and mass is in kilograms. If you wanted to use (for example) force in pounds, distance in furlongs, and mass in slugs, then you'd have to use a different value of G.

or if you used furlong slug and hour, then G = 6.67428 × 10⁻¹¹ would have to be multiplied by (metre/furlong)³/(kilogram/slug)(second/hour)²

 

[robertroman10]

so if I have two objects, one 100 kilograms and another 75, they are both 1 meter by one meter by one meter and are spaced 5 meters from each other what is the gravitational pull/attraction between them

 

[tiny-tim]

apply the formula! …

force = Gm₁m₂/r²​

 

[robertroman10]

DrGreg, jtbell, and tiny tim. Thank you so much I think I am starting to understand this! hopefully tonight will be my first calculations using Newtons universal gravitational constant! thanks so much everyone

 

[robertroman10]

So (just to make shure I am correct) I would:
1) find all variables
2) multiply the mass of the two objects I am using/comparing
3) divide this by the distance squared
4) multiply this by 6.67428*10^-11

is this right?

 

[DrGreg]

So (just to make shure I am correct) I would:
1) find all variables
2) multiply the mass of the two objects I am using/comparing
3) divide this by the distance squared
4) multiply this by 6.67428*10^-11

is this right?

Yes (using the units we've already agreed)

 

[FtlIsAwesome]

so if ... they are both 1 meter by one meter by one meter

To clarify, the sizes of the objects don't matter. Only their mass.
If you had two objects of equal mass, but different sizes, you'd still treat them the same in the equation.

So (just to make shure I am correct) I would:
1) find all variables
2) multiply the mass of the two objects I am using/comparing
3) divide this by the distance squared
4) multiply this by 6.67428*10^-11

is this right?

Correct.


Note that using ordinary masses, such as 10 kg and 5 kg, will result in a very small number. Gravity is weak, this is why we don't see household objects attracted to each other.



As an example, try calculating the force between Earth and the Moon.

Mass of Earth (kg): 6x10²⁴
Mass of Moon (kg): 7.3x10²²
Average distance between them (meters): 384,000,000

Answer below. Move mouse over it to see.

Spoiler

I get an answer of 2x10²⁰ N

 

[robertroman10]

Yes I got it! thanks everyone again for the help! soo excited i might just go measure the amount of Newtons I am putting on objects in my room because I am so bored:} thanks again!

 

[jtbell]

or if you used furlong slug and hour, then G = 6.67428 × 10⁻¹¹ would have to be multiplied by (metre/furlong)³/(kilogram/slug)(second/hour)²

Of course, the time unit that best matches with furlongs is fortnights, not hours.

The FFF system of units

 

[FtlIsAwesome]

Yes I got it! thanks everyone again for the help! soo excited i might just go measure the amount of Newtons I am putting on objects in my room because I am so bored:} thanks again!

Glad to be of assistance.

Of course, the time unit that best matches with furlongs is fortnights, not hours.

The FFF system of units

I learn new stuff everyday.

 

[tiny-tim]

Of course, the time unit that best matches with furlongs is fortnights, not hours.

The FFF system of units

oooh, thanks, jt … i'll definitely use fortnights in future! ​

i can't find any of the other units in the fff system …

lemmee see …

unit of angle: the farthing (= 90°, course )

unit of temperature: the fahrenheit

unit of charge: the flintstone

unit of energy: the sweetheart

unit of power: the fish! ​

any ideas for the unit of force?

 

[erok81]

To clarify, the sizes of the objects don't matter. Only their mass.

True the size doesn't matter, but where you find r from those masses does. Using from the surface or the center will give two different answers. I'm sure you know that, but maybe the the OP doesn't. Since distance and radius were both thrown around there may be some confusion.

To the OP, in this case you want to measure your r from the centers of each object you are using.

 

[FtlIsAwesome]

any ideas for the unit of force?

The falcon.

 

[FtlIsAwesome]

@robertroman10


In some cases people use kilometers instead of meters.

If this is so, then G will be:
6.67428x10^-20
Its units are N km²/kg²

When discussing the solar system, it is common list distances and other properties in kilometers. So check to make sure you haven't mixed meters and kilometers.

You will still get the same answer in Newtons.


This is the previous example using km.

Mass of Earth (kg): 6x10²⁴
Mass of Moon (kg): 7.3x10²²
Average distance between them (km): 384,000
Gravitational constant (N km²/kg²): 6.67428x10^-20

I get the exact same result of 2x10²⁰ N