GRAVITATION BYU 4

Solution 4

Analysis:

We are launching a satellite of mass $m$ from the surface of the Earth (Radius $R$) to a circular orbit of radius $r$.

• Initial State (Surface): $E_i = U_i + K_i = -\frac{GMm}{R} + 0$ (ignoring Earth’s rotation).
• Final State (Orbit $r$): Velocity $v = \sqrt{\frac{GM}{r}}$. Kinetic Energy $K = \frac{GMm}{2r}$. Potential Energy $U_f = -\frac{GMm}{r}$. Total Energy $E_f = -\frac{GMm}{2r}$.

Work Done ($W$):

The work done by the rocket is the change in total energy:

$$ W = E_f – E_i = -\frac{GMm}{2r} – \left( -\frac{GMm}{R} \right) = \frac{GMm}{R} – \frac{GMm}{2r} $$

We need a relationship between Kinetic Energy ($K$) and Work ($W$).

We know $K = \frac{GMm}{2r}$. Substituting this into the expression for $W$:

$$ W = \frac{GMm}{R} – K $$

Using $GMm/R = mgR$ (since $g = GM/R^2$):

$$ W = mgR – K \quad \Rightarrow \quad K = -W + mgR $$

Graph Limits:

1. Lowest Orbit ($r=R$): $$ K_{max} = \frac{mgR}{2} $$ $$ W = mgR – \frac{mgR}{2} = \frac{mgR}{2} $$ Point: $(mgR/2, mgR/2)$
2. Highest Orbit ($r \to \infty$): $$ K_{min} \to 0 $$ $$ W = mgR – 0 = mgR $$ Point: $(mgR, 0)$

The graph is a straight line with a negative slope (-1) strictly between these domains.