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: Write the conservation of mechanical energy equation.
At the earth's surface, the projectile has kinetic energy and gravitational potential energy.
At the maximum height, the projectile momentarily comes to rest, so kinetic energy becomes zero.
Using conservation of energy,
\[ \frac12 mV^2-\frac{GMm}{R} = -\frac{GMm}{R+h} \] where,
\[ V=\alpha V_E \] \[ h=800 \, km \] \[ R=6400 \, km \]

Step 2: Use the escape velocity relation.
Escape velocity is given by
\[ V_E=\sqrt{\frac{2GM}{R}} \] Squaring both sides,
\[ V_E^2=\frac{2GM}{R} \] Substitute \(V=\alpha V_E\) into the energy equation,
\[ \frac12 m\alpha^2V_E^2-\frac{GMm}{R} = -\frac{GMm}{R+h} \] Now replace \(V_E^2\) by \(\dfrac{2GM}{R}\),
\[ \frac12 m\alpha^2\left(\frac{2GM}{R}\right)-\frac{GMm}{R} = -\frac{GMm}{R+h} \] \[ \frac{GMm}{R}\alpha^2-\frac{GMm}{R} = -\frac{GMm}{R+h} \]

Step 3: Simplify the equation.
Divide throughout by \(GMm\),
\[ \frac{\alpha^2}{R}-\frac1R = -\frac1{R+h} \] \[ \frac{\alpha^2-1}{R} = -\frac1{R+h} \] Multiply both sides by \(R\),
\[ \alpha^2-1 = -\frac{R}{R+h} \] \[ \alpha^2 = 1-\frac{R}{R+h} \] \[ \alpha^2 = \frac{h}{R+h} \]

Step 4: Substitute the numerical values.
\[ \alpha^2 = \frac{800}{6400+800} \] \[ = \frac{800}{7200} \] \[ = \frac19 \] Taking square root,
\[ \alpha=\frac13 \]

Step 5: Final conclusion.
Hence, the value of the constant is
\[ \boxed{\frac13} \]