A projectile is fired straight upward from the Earth's surface at the South Pole with an initial speed equal to one third the escape speed. (The radius of the Earth is 6.38 106 m.)

(a) Ignoring air resistance, determine how far from the center of the Earth the projectile travels before stopping momentarily.

(b) What is the altitude of the projectile at this instant?

Question

A projectile is fired straight upward from the Earth's surface at the South Pole with an initial speed equal to one third the escape speed. (The radius of the Earth is 6.38  106 m.)

(a) Ignoring air resistance, determine how far from the center of the Earth the projectile travels before stopping momentarily.

(b) What is the altitude of the projectile at this instant?

anonymous · Answer

I'm having trouble figuring out how to do this.  The textbook gives the equation for escape velocity but that's already given.  Escape velocity = V/3. It seems like I should use a Kinematics equation for this using deceleration due to gravity.  I'm still missing something.

anonymous · Answer

It will go so high that gravity acceleration is not a constant.
Use formula for potential energy due to Earth gravity; then
set initial kinetic energy, (1/2) m v^2 equal to the final
potential energy and determine r, the distance from the center of the Earth. Subtract radius of Earth to get "altitude."

anonymous · Answer

Kinetic and Potential Energy are my weakest points.  Guess I'll have to review them.  Thanks.

anonymous · Answer

You are welcome.

irishboy123 · Answer

$$F = \frac{ GMm }{ r^{2} }$$

Are you familiar with this - Newton, Gravitational force?  If so read on.

Work = force * distance so in this case the potential energy that is converted from the initial kinectic energy of the projectile is given by:

$$PE = \int\limits_{r1}^{r2}\frac{ GMm }{ r^{2} }dr$$ 

$$= \left[ \frac{ -GMm}{ r }  \right]$$

or $$GMm(\frac{ 1 }{ r1 } - \frac{ 1 }{ r2 })$$

as described above equate the gain in PE with the loss in Kinetic energy.