Question

A projectile is shot directly away from Earth's surface. Neglect the rotation of Earth.
(a) As a multiple of Earth's radius RE, what is the radial distance a projectile reaches if its initial speed is one-fifth of the escape speed from Earth?
____ times RE

(b) As a multiple of Earth's radius RE, what is the radial distance a projectile reaches if its initial kinetic energy is one-fourth of the kinetic energy required to escape Earth?
_____ timesRE

(c) What is the least initial mechanical energy required at launch if the projectile is to escape Earth?


These problems all can be solved by using the Conservation of Energy relationship
Total Energy = KE + PE =
(1/2) m V^2 - GMm/R = constant
wher M is the mass of the earth, m is the mass of the projectile and R is the distance from the center of the Earth.
The "escape speed" Ve is the launch velocity that allows V to be zero when R = infinity. Thus
(1/2) m Ve^2 - G M m/Re = constant

Ve = sqrt (2 G M/Re)
Here is how to do (a):
If V = (1/5) Ve at R = Re, then
(1/2) m V^2 - G M m/R = constant
= (1/50) m Ve^2 - G M m/Re
When V = 0 (maximum value of R),
(1/50) Ve^2 = GM [(1/Re - (1/R)]
(1/25) G M/Re = G M [(1/Re - (1/R)]

GM/R = (24/25)(GM/Re)
R/Re = 25/24

Proceeds imilarly for (b)
For (c), the least mechanical energy (including potential energy), is zero, since potential energy is defined as zero at infinite distance, as is almost always done in inverse-square law situations. This is the case when PE = - G M m/R


Your explanaion is very clear to me. Thank you.

Answer 1

Sorry, your explanation made too many logical leaps. How did you get from 1/5 to 1/50??? It's garbage like this that makes you want to never see physics problems again.

Answer 2

Amen Max!

Answer 3

1/2(1/5Ve)^2 = 1/2 * 1/25 * Ve^2 = 1/50Ve^2

Answer 4

why is it 1/50??

Answer 5

Because (1/5Ve)^2 = 1/25Ve^2 You then multiply the 1/25 by the 1/2 to get 1/50.