A non-rotating spherical planet with no atmosphere has a mass M and radius R. A particle is fired...


A non-rotating spherical planet with no atmosphere has a mass {eq}M {/eq} and radius {eq}R {/eq}. A particle is fired off from the surface with a speed equal to {eq}\frac{3}{4} {/eq} the escape speed. Calculate the farthest distance it reaches (measured from the center of the planet) if it is fired tangentially.

Satellite Motion and Orbital Velocity:

An object launched from a given celestial body will orbit that body along an elliptical orbit if its launch velocity is smaller than the escape velocity. In the latter case, an object is not bound and can go to infinity. The motion along a closed bound orbit obeys conservation laws.

Answer and Explanation:

We will apply the law of conservation of energy and angular momentum to this problem.

{eq}\begin{align*} &m v_0 R = m v_a r_a &\text (1) \end{align*} {/eq}

This is the law of conservation of angular momentum for the particle;


  • {eq}m {/eq} is the mass of the particle;
  • {eq}v_0 = \dfrac {3}{4} \sqrt{\dfrac {2GM}{R^2}} {/eq} is the launch velocity;
  • {eq}v_a {/eq} is the speed of the particle at the point of farthest separation from the planet;
  • {eq}r_a {/eq} is the distance to the aphelion;

At the same time, the total energy is conserved too:

{eq}\begin{align*} &\dfrac {mv^2_0}{2} - \dfrac {G M m}{R} = \dfrac {mv^2_a}{2} - \dfrac {GMm}{r_a} &\text (2) \end{align*} {/eq}

Eliminating the velocity in aphelion, we obtain:

{eq}\dfrac {v^2_0 R^2}{2 r^2_a} - \dfrac {GM}{r_a} = \dfrac {v^2_0}{2} - \dfrac {GM}{R} {/eq}

Substituting the expression for {eq}v_0 {/eq}, we come up with the following equation:

{eq}\dfrac {9}{16} \cdot \dfrac {R}{r^2_a} - \dfrac {1}{r_a} + \dfrac {7}{16 R} = 0 {/eq}

Solving, we get:

{eq}r_a = \boxed{(8 + \sqrt{55})R} {/eq}

Learn more about this topic:

Stable Orbital Motion of a Satellite: Physics Lab

from Physics: High School

Chapter 8 / Lesson 16

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