An object of some mass, m, is accelerated upwards....

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An object of mass m is accelerated upwards from rest, and the discussion revolves around deriving the formula for the universal force of gravitation, FG = (1.6x10^15) x (m/x^2t^4). Participants express confusion about the problem's validity, particularly regarding the gravitational force's dependence on distance to the Earth's center rather than just height above the surface. One user suggests plugging in values for speed and time to test the formula's accuracy, while another reveals that a typo from the teacher led to the confusion. Ultimately, the consensus is that the formula cannot be correct as presented. The discussion highlights the importance of verifying problem statements in physics assignments.
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Homework Statement


an object of some mass, m, is accelerated upwards (from rest) off the surface off the Earth for some unknown time, t. At the end of this time, the object is traveling at some new speed, x. Derive the following formula. That is, show that no matter how much time the rocket travels for, as long as you know the speed of it in the end, the equation for the universal force of gravitation between the object and the Earth is given by:
FG = (1.6x10^15) x (m/x^2t^4)

Homework Equations


FG = (1.6x10^15) x (m/x^2t^4)

The Attempt at a Solution


I have attached a photo of my work, to the point that I got stuck on.
20180603_173848.jpg
 

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This problem doesn't quite make sense to me. Where did you get it?

(Note that the gravitational force depends on the distance to the center of the earth, not just on the distance above the surface.)
 
Doc Al said:
This problem doesn't quite make sense to me. Where did you get it?

(Note that the gravitational force depends on the distance to the center of the earth, not just on the distance above the surface.)

This is question I have been stuck on, in my weekly assignment. May you explain the part of the problem that doesn't make sense.
 
Vraj Patel said:
May you explain the part of the problem that doesn't make sense.
Plug in a few values for speed and time. If the distance traveled in that time is small, the gravitational force should equal mg. Does it?
 
Doc Al said:
Plug in a few values for speed and time. If the distance traveled in that time is small, the gravitational force should equal mg. Does it?

would you like me to plug it into the FG = (1.6x10^15) x (m/x^2t^4) formula.
 
Vraj Patel said:
would you like me to plug it into the FG = (1.6x10^15) x (m/x^2t^4) formula.
Yes.
 
Doc Al said:
Yes.
I am sorry for using your time, but my teacher made a typo, and I figured out the problem.
 
Vraj Patel said:
I am sorry for using your time, but my teacher made a typo, and I figured out the problem.
That's ok - it happens. (Hope I am not being presumptuous answering for Doc Al.)
 
Vraj Patel said:
I am sorry for using your time, but my teacher made a typo, and I figured out the problem.
No problem at all. I was leading you to plug in some values that would convince you (and your teacher!) that the given formula cannot possibly be correct. But no harm done. I'm glad your teacher found the typo.

haruspex said:
(Hope I am not being presumptuous answering for Doc Al.)
Not at all. :smile: I agree with you.
 
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