Using the gravitational formula

[462chevelle]

Homework Statement

if I was wanting to calculate the force of gravity at certain points relative to the Earth's surface. what number would I plug in the the constant of G in the formula. so far I have some terrible numbers and I want to make sure my G is correct. which I don't think it is. this isn't for homework. just for fun but I figured this was the best place to put it.

Homework Equations

Fg=G(m2m1/r^2)

The Attempt at a Solution

here is what I am plugging in
G = 1.2 x 10^-4 http://en.wikipedia.org/wiki/Gravitational_constant
that's where I got that. but I have the feeling ill need to use the other numbers. but I don't understand what that (80) is in there for
r=6371 km
m2=5.972 x 10^24
m1=90 kg
the answer I got was 1589016271417625
the number I was hoping for was. appx 9.8 m/s^2
then once I know I am doing it right I want to check and the relationship between being deeper and deeper into the Earth's surface and gravity.
thanks,
Lonnie

 

[462chevelle]

if I just plug in 6.67384 x 10^-11 I still get crap numbers.

 

[DeIdeal]

1.2 \times 10^{-4} is, as stated on the Wiki page, the standard uncertainty of the gravitational constant, the value of the constant itself is on the line above it, G\approx 6.67\times 10^{-11} \;\mathrm{m}^3 \mathrm{kg}^{-1} \mathrm{s}^{-2}. Remember to check your units, r=6371 \;\mathrm{km} =6371\times 10^3 \;\mathrm{m}= 6371000 \;\mathrm{m}.

You won't be getting the value you were hoping for, as you're calculating the force, not the acceleration due to gravity (which has the value of approx. 9.8 ms^-2 near the surface of the earth). Try to use F=ma to get the acceleration :)

 

[462chevelle]

oh ya. forgot about converting the units.
excuse my math skills but how is that formula computed? do you mind helping explain the exponents on the units.

 

[DeIdeal]

excuse my math skills but how is that formula computed?

Hm? Can't you just plug the values you now have into a calculator? Or are you looking for a derivation of the Newton's law of gravtitation, ie F=G\frac{m_1 m_2}{r^2}?

do you mind helping explain the exponents on the units.

Are you asking what they mean or why G has the units it does? If the latter: F, force, has the units [F]=\mathrm{N}=\mathrm{kg} \frac{\mathrm{m}}{\mathrm{s}^{2}}, mass has the units [m]=\mathrm{kg} and distance has the units [r]=\mathrm{m}.

F_G=G\frac{m_1 m_2}{r^2}\implies [G]=\frac{[F] [r]^2}{[m_1] [m_2]}=\frac{\mathrm{kgms}^{-2}\mathrm{m}^2}{\mathrm{kg}^2}=\mathrm{kg}^{1-2}\mathrm{m}^{2+1}\mathrm{s}^{-2}=\mathrm{m}^3 \mathrm{kg}^{-1} \mathrm{s}^{-2}

If the former, well, I don't really know what to say, that's just the way units work. When we, for example, multiply two physical quantities together, the unit of the resulting quantity is the product of the units of the quantities we multiplied.

Was that helpful at all?

 

[462chevelle]

do I put 6.67x10^-11 in the calc or what do I do something with those units

 

[DeIdeal]

You have to be consistent; when you put 6.67×10^-11 in your calculator, you have to have your other values as meters and kilograms, in order to get Newtons as a result but as long as you do that, everything should work.

 

[462chevelle]

just plugging it in I get 883

 

[462chevelle]

awesome that is perfect. thank you!

 

[DeIdeal]

Yes, that's about the magnitude of the gravitational force that is exerted on an object with a mass of 90 kilograms. If you want to get the acceleration due to gravity, you divide by 90 kg as per F=ma. The result should be close to the value you mentioned earlier.

EDIT: Glad I could help :]