Question

If a projectile is launched vertically from the Earth with a speed equal to the escape speed, how high above the earth's surface is it when its speed is half the escape velocity

.5mv^2 -(GMm)/r= .5m(v/2)^2-(GMm)/r

after doing the math on the right side I'm getting 2.08*10^5 using the radius of the earth
then after subtracting the half the escape velocity term I get -1.55*10^7. cross multiply r then divide by -1.55*10^7 to get 2.57*10^7 my book says this is the right answer but we didn't subtract the radius of the earth so how is that right?????

Answer 1

escape speed: Vo^2=2GM/R
in point that V=Vo/2 we have
1/2*m*Vo^2/4=G*M*m/(R+h)
.......
h=3R
h=3*6,400*10^3
h=19,200*10^3
your book mast calc the r =(R+h) and not the h

Answer 2

You must distinguish r on the left side of the equation and r on the right side
\[(mv _{e}^{2})/2-GMm /R _{Earth }=(m(v _{e}/2)^{2})/2-GMm /R _{unknown }\]
then you get

Answer 3

\[R _{unknown}=4 \times R _{Earth}\]

Answer 4

@Awstinf

Answer 5

So how do I know when I subtract and when I don't then??

Answer 6

I'm still confused why i don't subtract though....

Answer 7

"The projectile is launched at 11.2 km/s from the surface of the Earth, but slows down as it converts its initial kinetic energy into gravitational potential energy.
Strategy: Set the mechanical energy at the launch equal to the mechanical energy when the speed is equal to the one-half the escape speed. Then determine the gravitational potential energy and therefore the distance such a projectile would be from the center of the Earth." straight from the solution manual

Answer 8

the way they did it it seems as though they didn't account for the distance from the surface of the earth.... but idk though

Answer 9

You know I wonder why you must subtract anything? All the distances in your equation are from the center of Earth. even the unknown one. So you don't have to subtract Re from the result you got.

Answer 10

but it asks from the surface of the earth not the center how did the equation account for the surface of the earth? is it because the first radius is similar to the h in mgh?

Answer 11

You have to subtract at the very end.

The answer required is height ABOVE earth's surface, so:

\(\large h=R_{unknown}-R_{Earth}\)

Answer 12

Yes you are right. It seems to me that your book contains a mistake.

Answer 13

THANK YOU GOD! I was like wtf is wrong with this damn book

Answer 14

/golfclap to all of you!