Gravitational Force

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Calculus
Section 2.5

Screenshot_20220516-084301_Samsung Notes.jpg
 
graph:

upload_2022-5-21_21-9-40.jpeg



The function is certainly continuous on r element [0, R) and r element (R,∞).
The only question is whether it is continuous at r = R.
Checking the limit from each side, we find that

upload_2022-5-21_21-7-48-gif.3146

R^-)( piecewise | (G M r)/R^3 | r<=R (G M)/r^2 | r>=R) = (G M)/R^2" style="-x-ignore: 1" data-paste-id="1">
upload_2022-5-21_21-8-23-gif.3147


R^+)( piecewise | (G M r)/R^3 | r<=R (G M)/r^2 | r>=R) = (G M)/R^2" style="-x-ignore: 1" data-paste-id="2">
Both limits are equal, and the value of F(r) is equal to the limit because of the r ≥ R
clause in the function definition. Therefore the function is continuous for any r > 0.

 

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graph:

View attachment 3148


The function is certainly continuous on r element [0, R) and r element (R,∞).
The only question is whether it is continuous at r = R.
Checking the limit from each side, we find that

upload_2022-5-21_21-7-48-gif.3146

R^-)( piecewise | (G M r)/R^3 | r<=R (G M)/r^2 | r>=R) = (G M)/R^2" style="-x-ignore: 1" data-paste-id="1">
upload_2022-5-21_21-8-23-gif.3147


R^+)( piecewise | (G M r)/R^3 | r<=R (G M)/r^2 | r>=R) = (G M)/R^2" style="-x-ignore: 1" data-paste-id="2">
Both limits are equal, and the value of F(r) is equal to the limit because of the r ≥ R
clause in the function definition. Therefore the function is continuous for any r > 0.

Thanks. Did you try uploading several images? If so, only one image is in display.
 
upload_2022-5-21_21-13-41-gif.3151


upload_2022-5-21_21-13-52-gif.3152

did you see these imgs

Now I can see the images. Thank you.

Note:

It is really humid in NYC. This morning the humidity was 80%. I don't think it has gone down. I do not have an air conditioner in my room. My fan is blowing hot air.

I took two showers today but to no avail. So, in light of this uncomfortable weather condition, I will not do the Intermediate Value Theorem tonight. I will leave it for another day.
 

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