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AP EAMCET · PHYSICS · Gravitation

An object is thrown vertically upwards from the surface of the earth with a velocity \(x\) times the escape velocity on the earth \((x < 1)\), then the maximum height to which its rises from the centre of the earth is (radius of earth is \(R\) )

  1. A \(R(1-x)^2\)
  2. B \(\frac{Rx^2}{\left(1-x^2\right)}\)
  3. C \(\frac{1-x^2}{R}\)
  4. D \(\frac{x^2}{1-R}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(\frac{Rx^2}{\left(1-x^2\right)}\)

Step-by-step Solution

Detailed explanation

(No option is matching.) Escape velocity is \(v_e=\sqrt{\frac{2 G M}{R}}\) Let object rises upto height \(h\), when it is thrown with speed \(x \cdot v_e\), then from conservation of energy, we get \( \frac{1}{2} m\left(x \cdot v_e\right)^2-\frac{G M m}{R}=0-\frac{G M m}{R+h} \)…