- #1
directory_NA
- 10
- 0
Homework Statement
A particle of mass m is dropped from a height h, which is not necessarily small compared with the radius of the earth. Show that if air resistance is neglected, the speed of the particle when it reaches the surface of the Earth is given by ##\sqrt {2gh}## ##\sqrt {\frac {R_E} {R_E + h}}## .
(##R_E## = radius of Earth)
Homework Equations
##KE = \frac 1 2##mv2
##PE = -G \frac {Mm} {r}##
##PE= mgh## (unsure if this one is relevant to the question)
The Attempt at a Solution
I started with the basic rule of conservation of energy and wrote out my equation:
##\frac {1} {2} mv^2_i - \frac {GMm} {R_E + h} = \frac {1} {2} mv^2_f - \frac {GMm} {R_E}##
Since it was dropped, thus, I went ahead and presumed ##v_i = 0## and also canceled out all the "m"'s. Giving the equation:
##- \frac {GM} {R_E + h} = \frac {1} {2} v^2_f - \frac {GM} {R_E}##
If I solve for ##v_f##, I would get the following:
##\sqrt {2GM[R_E - (R_E + h)]}##
... which consequently looks nothing like the desired equation.
I'm not sure if I'm on the right track or something is going wrong within my derivation so I would like some advice regarding how to approach this problem and what would lead me on the right track.