(adsbygoogle = window.adsbygoogle || []).push({}); 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-third of the escape speed from Earth?

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

U = -GMm/r

K = 1/2mv^2

Radius of earth = 6.38e6

Mass of earth = 5.98e24

(a) Escape speed of earth = 11.2 km/s

=11200 m/s

U1 + K1 = U2

-GMm/r + 1/2mv^2 = -GMm/(r+h)

m cancels out

11200/3 = 3733

(-6.67e-11 * 5.98e24) / 6.38e6 + 1/2 * 3733^2 = (-6.67e-11 * 5.98e24) / (6.38e6 + h)

-55550537.32 = -3.98866e14 / (6.38e6 + h)

h = 800236.5781

h / Re = 800236.5781 / 6.38e6 = 0.125 times the radius of the earth

(b)

U1 + K1 = U2

-GMm/r + 1/2mv^2 = -GMm/(r+h)

but it asks for 2/5 the KE, so

-GMm/r + (2/5)1/2mv^2 = -GMm/(r+h)

m cancels out

-GM/r + 1/5 v^2 = -GM / (r+h)

Plugged in the same numbers as part (a), except for v, which is 11200 m/s

I got h, divided by Radius of earth (6.38e6) and get 0.670 times radius of earth

Did I make any mistakes? Because the answers are wrong, apparently

**Physics Forums - The Fusion of Science and Community**

# Gravitational Potential Energy of a projectile

Know someone interested in this topic? Share a link to this question via email,
Google+,
Twitter, or
Facebook

Have something to add?

- Similar discussions for: Gravitational Potential Energy of a projectile

Loading...

**Physics Forums - The Fusion of Science and Community**