Question

The escape velocity from the earth is \(v_e\). If an object is now launched from the center of Earth where a tunnel has been dug, the escape velocity from that position is:

• A. \(v_e \sqrt{\frac{3}{2}}\)
• B. \frac{2v_e}{\sqrt{3}}
• C. \(v_e / \sqrt{2}\)
• D. \(v_e \sqrt{2}\)

Hint:

The potential at the center of a uniform solid sphere is always exactly \(1.5\) times (or \(\frac{3}{2}\)) the potential at its surface. Consequently, the square of the escape velocity from the center scales directly by this same factor.

Answer 1

The Correct Option is A

Concept: Escape velocity (\(v\)) from any point is the minimum speed required for a particle to escape the gravitational field of a massive body completely (reaching infinity with zero kinetic energy). It is derived using the law of conservation of mechanical energy: \[ E_i = E_f \quad \Rightarrow \quad K_i + U_i = K_f + U_f \] At infinity, both potential energy and kinetic energy are zero (\(U_f = 0, K_f = 0\)). Thus: \[ \frac{1}{2}m v^2 + U_i = 0 \quad \Rightarrow \quad v = \sqrt{\frac{-2U_i}{m}} \] Step 1: Relating the standard surface escape velocity \(v_e\).
The gravitational potential energy of a mass \(m\) on the surface of a solid uniform sphere (Earth) of mass \(M\) and radius \(R\) is: \[ U_{\text{surface}} = -\frac{GMM}{R} \] Substituting this into the energy balance gives the standard surface escape velocity: \[ \frac{1}{2}m v_e^2 - \frac{GMm}{R} = 0 \quad \Rightarrow \quad v_e = \sqrt{\frac{2GM}{R}} \quad \cdots (1) \]

Step 2: Finding the potential energy at the center of the Earth.
The gravitational potential \(V\) inside a uniform solid sphere at a distance \(r\) from the center is given by the formula: \[ V(r) = -\frac{GM}{2R^3}(3R^2 - r^2) \] At the center of the Earth, \(r = 0\). Substituting this value yields the potential at the center: \[ V_{\text{center}} = -\frac{3GM}{2R} \] Therefore, the gravitational potential energy \(U_{\text{center}}\) of a mass \(m\) at the center is: \[ U_{\text{center}} = m \cdot V_{\text{center}} = -\frac{3GMm}{2R} \]

Step 3: Calculating the new escape velocity \(v'\) from the center.
Using the conservation of energy from the center to infinity: \[ \frac{1}{2}m (v')^2 + U_{\text{center}} = 0 \quad \Rightarrow \quad \frac{1}{2}m (v')^2 - \frac{3GMm}{2R} = 0 \] \[ (v')^2 = \frac{3GM}{R} \quad \Rightarrow \quad v' = \sqrt{\frac{3GM}{R}} \quad \cdots (2) \]

Step 4: Expressing \(v'\) in terms of \(v_e\).
Divide equation (2) by equation (1): \[ \frac{v'}{v_e} = \frac{\sqrt{\frac{3GM}{R}}}{\sqrt{\frac{2GM}{R}}} = \sqrt{\frac{3}{2}} \] \[ v' = v_e \sqrt{\frac{3}{2}} \]