What is this one's definite integral ?

[Calton]

Main Question or Discussion Point

This is too far behind me ( time-wise ), and I've gotten different answers from virtually every different online integral calculator . What is the integral of : ( 2.12068539072 x 10^-11 ) / x^2 from 6,378,100 to 1,500,000,000 ? Please only help if you really know your maths well .

 

[Drakkith]

I believe the top of the fraction can be brought to the outside of the integral, leaving you with 1/x² on the inside. Now it's a near-trivial integral with some large limits of integration. Do you know the antiderivative of 1/x²?

 

[Calton]

I just reposted this question cos I thought it didn't take. The integration - as far as I can determine comes to just : -1/x So then I think it gets worked out as
1/b - 1/a where they are ( respectively ) 1500000000 & 6378100 . The answer I get comes to : 1.5678650e-7 - Which I think is wrong.
I'm trying to calculate the average force on an O2 molecule 1,500,000,000 meters out from Earth - coming to ground. It SHOULD also be calculated as from infinity as well. Then I think it should cross check (when its arrival speed is calculated ) with the escape velocity for Earth ( 11.2 Km/s ) .

 

[Drakkith]

I'm about to head to bed, but I'll try to remember to come back to this after I wake up tomorrow.

 

[Erland]

I get 3.31081057847891 *10^-18 with GNU Octave.

 

[Delta2]

I think what you actually calculate with this integral is the work of the gravity force from 6378100 to 1500000000, you have to divide the value of the integral to the (1500000000-6378100) to get the average force.

 

[fresh_42]

I get 3.31081057847891 *10^-18 with GNU Octave.

The ordinary Windows calculator gives the same result.

 

[Calton]

I get 3.31081057847891 *10^-18 with GNU Octave.

Thank you Erland. - Yes, - I found, - many hours later, the same number that YOU have come up with. I'm going to have a look at that "GNU Octave" you've mentioned. - Once again, - many thanks.

 

[Calton]

I think what you actually calculate with this integral is the work of the gravity force from 6378100 to 1500000000, you have to divide the value of the integral to the (1500000000-6378100) to get the average force.

- Thank you Delta for you r effort too ! - Also : - No. The answer is the Work done. Then negating that answer, gives the Potential Energy accumulated by that
movement ( out to the greater distance ). - Once again many thanks.

 

[Calton]

Dear Delta,
- I must apologize for my reply to you. I was driving along in the car the next day, and realized how dumb what I wrote was. - You were basically correct, ( except you must negate like I said ) - then dividing by the distance gives the average force, - like you said. - Thanks again.