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:
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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.