Question

Consider an airless, non-rotating planet of mass M and radius R. An electromagnetic launcher standing on the surface of this planet shoots a projectile with initial velocity v0 directed straight up. Unfortunately, due to some error, v0 is less than the planet’s escape velocity ve; specifically, v0 = 0.701 ve. Unable to escape the planet’s gravitational pull, the projectile rises to a maximal height h above the ground, then falls back to the ground. Calculate the ratio h R of the projectile’s maximum height to the planet’s radius.

Answer 1

Answer 2

 1. Recall Key Formulas:

• Escape velocity ($v_e$):

  $$v_e=\sqrt{\frac{2GM}{R}}$$

• Given initial velocity ($v_0$):

  $$v_0=0.701v_e=0.701\sqrt{\frac{2GM}{R}}$$

2. Energy Conservation Approach:

At launch (surface) and max height $h$, total mechanical energy is conserved:

$$\text{Initial Energy}=\text{Energy at Max Height}$$$$\frac{1}{2}m{v}_0^2-\frac{GMm}{R}=-\frac{GMm}{R+h}$$

3. Simplify the Equation:

Divide both sides by $m$ and substitute $v_0$:

$$\frac{1}{2}(0.701{)}^2\left(\frac{2GM}{R}\right)-\frac{GM}{R}=-\frac{GM}{R+h}$$$$(0.701{)}^2\frac{GM}{R}-\frac{GM}{R}=-\frac{GM}{R+h}$$

Factor out $\frac{GM}{R}$:

$$((0.701{)}^2-1)\frac{GM}{R}=-\frac{GM}{R+h}$$$$(0.491-1)\frac{GM}{R}=-\frac{GM}{R+h}$$$$-0.509\frac{GM}{R}=-\frac{GM}{R+h}$$

Cancel negatives and $GM$:

$$0.509\frac{1}{R}=\frac{1}{R+h}$$

4. Solve for $h$:

$$R+h=\frac{R}{0.509}$$$$h=R(\frac{1}{0.509}-1)=R(1.964-1)=0.964R$$

5. Final Ratio:

$$\frac{h}{R}=0.964\approx 0.96$$