Gravitational Potential Energy of a projectile

[lu6cifer]

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

 

[millitiz]

Well, at least the eq looks right for me. The only thing I could think of, is that h is counted also from the center of earth. So instead of .125, h is 1.125?

 

[nlightner]

.

No, your calculations and approach seem correct. It is possible that there may be a rounding error or a mistake in the given answers. It is always a good idea to double check your calculations and make sure you are using the correct units for all quantities. You may also want to ask for clarification or double check with your teacher or textbook to ensure you have the correct answers.