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Gravitational Potential Energy of a projectile

  • Thread starter lu6cifer
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  • #1
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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
 

Answers and Replies

  • #2
79
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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?
 

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