# Projectile Launch - Earth's Radius

1. Mar 21, 2005

### vworange

The answer that the book gives for the question:
What is the launch speed of a projectile that rises above the Earth to an altitude equal to the Earth's radius? Ans. 7.91 km/s

Now I have the question:
What is the launch speed of a projectile that rises above the Earth to an altitude equal to three times the Earth's radius?

So I use:
$$v_{f} = \sqrt(((2GM_{e})/(4R_{e}))$$

Note:
G = 6.67x10^-11
Me = 5.97x10^34 kg
Re = 6.37x10^6 m

The resulting answer I get is 5.59 km/s. That equation works for an altitude equal to the earth's radius, but not for my problem. Why not? What am I doing wrong?

2. Mar 21, 2005

### Nylex

Possibly using 4R_{e} instead of 3R_{e}?

3. Mar 21, 2005

### Andrew Mason

The correct energy equation is:

$$KE = \Delta PE = GMm(\frac{1}{R_{i}}-\frac{1}{R_{f}})$$

$$v_i= \sqrt{2GM(\frac{1}{R_e}-\frac{1}{4R_e})}$$

BTW, the mass of the earth is 5.97x1024 kg.

AM