Question

A projectile is thrown straight upward from the earth's surface with an initial speed \(V=\alpha V_E\), where \(\alpha\) is a constant and \(V_E\) is the escape speed. The projectile travels upto a height \(800 \, km\) from earth's surface before it comes to rest. The value of the constant \(\alpha\) is
\[ (\text{Radius of the earth} = 6400 \, km) \]

• A. \(\dfrac{1}{3}\)
• B. \(\dfrac{1}{2}\)
• C. \(\dfrac{2}{3}\)
• D. \(\dfrac{3}{4}\)

Hint:

For projectile motion under gravity at large heights, use conservation of mechanical energy with gravitational potential energy: \[ U=-\frac{GMm}{r} \] instead of the near-earth formula \(mgh\).

Answer 1

The Correct Option is A

Step 1: Use energy conservation in the gravitational field.
At launch the projectile has kinetic energy plus gravitational potential energy at the surface, and at the top it is momentarily at rest with only potential energy. So \[ \frac12 mV^2-\frac{GMm}{R}=-\frac{GMm}{R+h} \]
Step 2: Bring in the escape speed.
Escape speed satisfies $V_E^2=\dfrac{2GM}{R}$, and the launch speed is $V=\alpha V_E$, so $V^2=\alpha^2 V_E^2=\alpha^2\dfrac{2GM}{R}$.
Step 3: Substitute into the energy equation.
The kinetic term becomes $\dfrac12 m\,\alpha^2\dfrac{2GM}{R}=\alpha^2\dfrac{GMm}{R}$, so $\alpha^2\dfrac{GMm}{R}-\dfrac{GMm}{R}=-\dfrac{GMm}{R+h}$.
Step 4: Cancel and simplify.
Dividing through by $GMm$, $\dfrac{\alpha^2-1}{R}=-\dfrac{1}{R+h}$, which rearranges to $\alpha^2=1-\dfrac{R}{R+h}=\dfrac{h}{R+h}$.
Step 5: Put in the numbers.
With $h=800$ km and $R=6400$ km, \[ \alpha^2=\frac{800}{6400+800}=\frac{800}{7200}=\frac19 \]
Step 6: Take the root and conclude.
So $\alpha=\dfrac13$.
\[ \boxed{\dfrac13} \]