What's the force (Nm) required to move an object (that weighs 1000kg) from earths surface, to the height of 1000km? (Earths radius is 6370km and earths mass is (5.98*10^24)kg)

Question

mos1635 · Answer

force in Nm ?? what is that ?
it seems like tork??

anonymous · Answer

I am not certain, all I know is that the final answer should be in Nm.
Moreover, I think you're supposed to use:$$F=(k(m1)*(m2))/r^2$$ k=6,67*10^(-11)Nm^2/kg^2

jamesj · Answer

Yes, Newton.meters are more commonly known as Joules and are the units for work.  In general, the amount of work done on an object is equal to its change in energy,
$$ W = \Delta E $$

The amount of work you have to do to move the object from the surface of the earth to 1000 km is the the change in the objects gravitational potential energy.

Hence what you have to do is find the gravitational PE for those two locations and take their difference,
$$ W = \Delta E = PE_{1000 \ km} - PE_{surface} $$

One last hint: if you don't already know the formula for the PE, work can be calculated explicitly be evaluating a line integral
$$ W = \int_A^B F \cdot d\ell $$

mos1635 · Answer

no pbjection to that but that is not force!!!!

fretje · Answer

@mos1635 : yes, the question should be: what is the work (Nm) required...

fretje · Answer

cannot use formula for potential energy and than subtract, g changes too much, must integrate, to be accurate.

anonymous · Answer

Indeed it was my faulth, translating the problem from Swedish --> English.
To integrate was way to go, indeed! :)

fretje · Answer

one can also be Buddhist, and just meditate on the object until it floats to required height

jamesj · Answer

Yes, integration is the way to go. If you're unsure of the theory, I recommend watching this excellent lecture: http://ocw.mit.edu/courses/physics/8-01-physics-i-classical-mechanics-fall-1999/video-lectures/lecture-11/

jamesj · Answer

@fretje, that explains why Tibet lead the world in physics and engineering

fretje · Answer

ok, so
$$W (work) = \int\limits_{r}^{r+1000} F.ds = \int\limits_{r}^{r+1000} \frac{G.1000kg.5,98.10^{\left( 24 \right)}kg.ds}{r^2}$$
where r is the distance between center of masses Mearth and M object.

$$W (work) = G.1000kg.5,98.10^\left( 24 \right)kg \int\limits_{6370000}^{7370000} \frac {ds}{r^2} $$

$$W (work) = G.1000kg.5,98.10^\left( 24 \right)kg \left[ \frac{-1}{r} \right]_{7370000m}^{6370000m}$$
where G = universal gravitation constant : 6,673×10-11 m3.kg-1.s-2

$$W (work) = 6,673.10^{-11}m ^{3}.kg ^{-1}.s ^{-2}.1000kg.5,98.10^{24} kg \left( \frac{-1}{6370000m} + \frac{1}{7370000m}\right)$$

which is
8,5GJ, a bit less than formula 
W = m.g.h = 1000kg.9,81m.s-2.1000000m = 9,81 MJ