Question reguarding gravitational pull

  • Thread starter Chewy0087
  • Start date
370
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Hey there, firstly i'll just say to start i've not studied general relativity at all, and the answer to my question may lie there, but i'll ask anyway;

Well I was thinking about the gravitational equation;

F (1,2) = m1m2G/d²

However putting the earth & a person in there gives an absolutely ridiculously high result, and yet we know for a fact that the pull of gravity on us is constant (10 ms ish), by just using really rough internet figures i get

(70 (mass of avg person) * 5.9 * 10^24 (earth mass) * G)/ 6377000^2 (dist from core - surface of the earth)

= 583 ish ms? A ridiculous answer

Any reason for this? I don't think i've gone much wrong, not enough to cause that :eek:

Thanks in advance for any help

Edit :JEEZ MEGA FAIL ON MY PART - F = MA, LAWLLLLL, diving by 70 gives a very good answer for gravity -

does this then mean that 9.8 isn't constant? And it is indeed dependant on mass, but it makes almost no difference whatsoever - sorry for failthread
 
A person with a mass of 70 kg has a weight close to 700 newtons on earth. That's the gravitational force exerted by the earth on the person. Your method should give you a number close to that expected answer.
 
G m_earth / (r_earth)^2

= (6.67 X 10^-11 N m^2/kg^2) ( 5.98 X 10^24 kg) / (6.37 X 10^6 m)^2

= 9.83 kg m/s^2
 
370
0
So....it's not constant? :P Didnt really get the 2nd post
 

DaveC426913

Gold Member
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(70 (mass of avg person) * 5.9 * 10^24 (earth mass) * G)/ 6377000^2 (dist from core - surface of the earth)
Where are you incorporating G, which equals 6.67*10^-11?
 
370
0
Yeah sorry i forgot to edit the original post - i did get it in the end, just forgot to divide by mass xP, got 9.75 ish which is awesome considering the mega-rounding i did with the wiki figures, it's just really nice to see these awesome equations work using them yourself! Thanks guyz
 

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