A satellite fired off a planet's surface

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SUMMARY

The discussion focuses on calculating the farthest distance a particle reaches from the center of a nonrotating spherical planet when fired tangentially at 3/4 of the escape speed. The relevant equations include gravitational potential energy (U = MmG/r) and the conservation of energy principle (E1 = E2). The initial velocity is determined to be v_o = 3/4sqrt(G2M/R), leading to the equation 9/32(G2M/R) = 2GM/r^3. The final expression for the distance from the center of the planet is derived as r = 32R/18.

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  • Understanding of gravitational potential energy and forces (U = MmG/r, F = -MmG/r^2)
  • Familiarity with conservation of energy principles in physics
  • Knowledge of basic calculus for solving integrals
  • Ability to manipulate algebraic expressions and solve for variables
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This discussion is beneficial for physics students, educators, and anyone interested in celestial mechanics and gravitational physics.

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Homework Statement


A nonrotating spherical planet with mass M and no atmosphere has radius R. A particle is fired off the surface at 3/4 the escape speed. Calculate the farthest distance it reaches from the center of the planet if it's fired of tangentially.


Homework Equations


l = r x p
U = MmG/r
F = -MmG/r^2
dKE = Fdr

The Attempt at a Solution


E1 = E2
1/2mv^2 - GMm/r = 0
1/2mv^2 = GMm/r
V(escape) = sqrt(2GM/r)
v_o = 3/4sqrt(G2M/R)

So I solved the initial velocity, but I'm confused as to what to do next. I tried this:
d1/2m(3/4sqrt(G2M/R))^2 = -MmG/r^2dr
and solving for r
so
9/32(G2M/R) = 2GM/r^3
9/32(1/R) = 1/r^3
r = cubedroot(1/(9/32*1/R)

Am I right?
 
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I'm sorry, the integral should be 18/(32R) = 1/r
r = 32R/18
 
Can someone please help me with this?
 

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