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 \(R\) )
- A \(R(1-x)^2\)
- B \(\frac{Rx^2}{\left(1-x^2\right)}\)
- C \(\frac{1-x^2}{R}\)
- D \(\frac{x^2}{1-R}\)
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 \( \frac{1}{2} m\left(x \cdot v_e\right)^2-\frac{G M m}{R}=0-\frac{G M m}{R+h} \)…
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