Who can tell me aprox. how much the earth weights and also the formula used to find it

Chris

hint the answer should look like "X.XX 10^XX"

 

What does the winner get?



Wiz

[link:www.clubfte.com/users/thewiz427/index.html|"Wiz's Place"]

 

[updated:LAST EDITED ON 28-Oct-02 AT 06:47 PM (EST)]The pride of knowing that they were the first person to answer the question correctly.

Chris

 

6.0 x 10^24 kg

you can use F=GmM/r^2

 

Compared to what?

It doesn't weigh a thing. It just floats out there in space.

Don't the objects on earth weigh a certain amount according the the density of the object and the density of the earth, which dictates the earths gravity? The denser the planet, the greater the gravity. And the denser the object the more it is affected by that gravity, am I right? I mean if the earth were to rest on the sun without burnning up how much would it weigh compared to lets say, Jupiter. What's the density of the sun and it's gravitional pull against the earth? What the density of the earth and how fast does it have to travel through space to maintain an orbit around the sun? Could this be part of the formula?

Actually I don't know what the heck I'm talking about.

This is way too complex for this little mind. Just pondering the question.

 

It was just 275 lbs lighter I jumped

 

6,000,000,000,000,000,000,000,000 kilograms is the weight of the mass . Make sense ?



Jeff

 

The method he used is by the summation of forces. Basically, the law of gravitation states that all matter in the universe is attracted to each other, as perscribed by the afore mentioned formula:

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

where:
F=net force due to gravitation
G=Universal gravitational constant (6.672*10^-11 N*m^2/Kg^2)
m1, m2 are the masses of the objects under investigation
r=the distance between the two objects

as you can see from the formula, great distances really make this force negligible, as twice a far away would make the force only 1/4 as great (inverse squares). To solve for the mass of the earth, you rearrange this equation, knowing that the net gravitational force that we feel on earth is the sum of all the forces of each and every particle on the earth pulling on us, which we commonly call gravity. I assumed a 1kg mass, which feels a force of 9.81 Newtons, at a distance from earth of (6,371,000 m), the average radius of the earth. Solving for the mass of the earth, I get 9.36*10^17 Kg. I've seen it listed in textbooks as 8.something, but I can't remember.

Density is the ratio of mass to volume of a body, so density does have an effect on gravitational pull, but ultimately only the mass of the body and the distance between the two will determine gravitational force.

'77 F100, 302 (the aftermarket Prodigy), C4
Cadet Second Lieutenant John F. Daly III
South Carolina Corps of Cadets, The Citadel
The TorqueKing

 

To the TorqueKing:

The formula is correct, but the numbers didn't come out right.

F := force of attraction between masses
G := universal gravitation constant
M := mass of the earth
m := mass of arbitrary 'test mass' at Earth's surface
r := radius of the earth

Newton's (I think) shell theorem allows to treat the earth as a point mass in this problem, since our 'test mass' lies on or outside the actual radius of the earth.

F == GMm/r^2 --> M == Fr^2/Gm == (F/m)r^2/G

But F/m is the gravitiational acceleration at Earth's surface for any mass m, which is g, the empirically determined constant.

g = 9.81 m/s^2
r = 6.371E6 m
G = 6.67E-11 Nm^2/kg^2

--> M == gr^2/G = 5.97E24 kg

(physics major)

 

I got the same equation, but when I substituted F/m for gravity (9.81m/s^2), my units got jacked up. That's because I wasn't paying attention to the fact that a Newton is a Kg*m/s^2. Got it, thanks for the correction, I got the same answer you did now. (Civil Engineering Major). TK

'77 F100, 302 (the aftermarket Prodigy), C4
Cadet Second Lieutenant John F. Daly III
South Carolina Corps of Cadets, The Citadel
The TorqueKing

 

Bonus question:

Who can tell me the minimum velocity with which an object would need to be launched from the earth, and it would never, ever come back? (Yes, such a velocity exists.)

Oh yeah, and how to calculate it using F == GMm/r^2?

 

8km per second. it would fall at 4.9m per second.

Chris

 

Okay, I got it! Solve the equation for gravity, once we substituted F/m, then the integral of gravity (acceleration) is velocity, then integrate the right side with respect to "R", as in the rate of change of position. My units came out m^2/s^2, so I took the square root to get:

-7906 m/s, in other words, to escape the earth's pull, this velocity must be exceeded. The negative sign indicates that it must be directly in the opposite direction that gravity acts in, in other words, the component opposite gravity must exceed that value.

Thanks for the challenge! TK


'77 F100, 302 (the aftermarket Prodigy), C4
Cadet Second Lieutenant John F. Daly III
South Carolina Corps of Cadets, The Citadel
The TorqueKing

 

Hate to say this but: You're both wrong.