MCQ
A certain object is projected vertically from the surface of the earth of radius $R$ with a velocity equal to half the escape velocity. The maximum height attained by the object will be
- ✓$R/3$
- B$2R$
- C$3R$
- D$6R$
✓
Answer
Correct option: A.
$R/3$
a
Let $\mathrm{h}$ be the maximum height reached by the particle. Potential energy on the surface of earth
Let $\mathrm{h}$ be the maximum height reached by the particle. Potential energy on the surface of earth
$=-\frac{\mathrm{GMm}}{\mathrm{R}}$
$PE$ at an altitude $\quad \mathrm{h}=-\frac{\mathrm{GMm}}{\mathrm{R}+\mathrm{h}}$
Escape velocity
$\mathrm{v}=\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}}$
The $KE$ of the object is,
$\frac{1}{2} \mathrm{m}\left(\frac{1}{2} \mathrm{v}\right)^{2}=\frac{1}{8} \mathrm{mv}^{2}$
$=\frac{1}{8} \mathrm{m} \times \frac{2 \mathrm{GM}}{\mathrm{R}}=\frac{\mathrm{GMm}}{4 \mathrm{R}}$
By Law of conservation of energy,
$\frac{\mathrm{GMm}}{\mathrm{R}}-\frac{\mathrm{GMm}}{\mathrm{R}+\mathrm{h}} \frac{\mathrm{GMm}}{4 \mathrm{R}}$
Solving, we get; $\mathrm{h}=\frac{\mathrm{R}}{3}$
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