Recent content by directory_NA

Thank you!

I see it now, that was sly. Thanks so much!

So the implication is that algebraically my equation is the same as the one in the question, but it is just organized in a different manner? And the only way to get to the question equation would be to fiddle with the algebra a bit more? Is it possible to give any hints regarding how to approach...

I managed to get ##v = \sqrt {2gR_E} \sqrt {(1-\frac {R_E} {R_E+h})}## However, now there seems to be an extra ##R_E## and an extra ##(1-)##

Let me take a look at it for a second...

##R_E##

Sorry, I'm not quite sure what you mean. I tried substituting ##\frac {GM} {r^2}## into the desired equation in the question, but it didn't seem to connect with what I had.

##\frac {GM} {r^2}##, if the m was taken out from ##mg = \frac {GMm} {r^2}##

Ah, now looking back I realize that the ##R_E - (R_E + h)## should be instead, ##\frac {1} {R_E} - \frac {1} {R_E + h }## Yet that doesn't quite account for why ##R_E## would be in the numerator and there is a ##gh## instead of ##GM##

May I ask where the problem arose? I'm having trouble finding the root of the error.

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}...

Hey, everyone. I've just joined this forum and probably will be coming here quite often. I'm a high school student in Canada and will mainly be seeking help here for questions regarding ap physics.