Question

A particle is projected vertically upwards from the surface of the earth (radius Re) with a speed equal to one forth
of escape velocity. What is the maximum height attained by it?
(A) 16/15 Re
(C) 4/15Re
(B) Data insufficient
(B) None of the above​

Answer 1

Answer:

Option D is the correct answer

Explanation:

From conservation of mechanical energy

$= > \frac{1}{2} m {v}^{2} = \frac{GMm}{ R_{e}} - \frac{GMm}{R}$

Where-:

R = Maximum distance from centre of the earth and

$= > v = \frac{1}{4} v_{e} \\ = > v = \frac{1}{4} \sqrt{ \frac{2GM}{R_{e}} } \\ = > \frac{1}{2} m \times \frac{1}{16} \times \frac{2GM}{R_{e}} = \frac{GMm}{ R_{e}} - \frac{GMm}{R} \\ = > \frac{1}{ \cancel2} m \times \frac{1}{16} \times \frac{ \cancel2GM}{R_{e}} = \frac{GMm}{ R_{e}} - \frac{GMm}{R} \\ = > R = \frac{16}{15} {R_{e}} \\ = > h = R - R_{e} \\ = > \boxed{h = \frac{R_{e}}{15} }$

Answer 2

The maximum height is $\dfrac{R_{e}}{15}$

(D) None of the above

Explanation:

Given that,

Speed $v=\dfrac{1}{4}v_{e}$

We need to calculate the radius of the earth

Using conservation of energy

$K.E=P.E-\dfrac{GMm}{R}$

$\dfrac{1}{2}mv^2=\dfrac{GMm}{R_{e}}-\dfrac{GMm}{R}$

Where, R = maximum height from the center of the earth

Put the value into the formula

$\dfrac{1}{2}\times\dfrac{1}{16}\times\dfrac{2GM}{R_{e}}=\dfrac{GM}{R_{e}}-\dfrac{GM}{R}$

$\dfrac{1}{16R_{e}}=\dfrac{1}{R_{e}}-\dfrac{1}{R}$

$R=\dfrac{16}{15}R_{e}$

We need to calculate the maximum height

Using formula of height

$h = R-R_{e}$

Put the value into the formula

$h=\dfrac{16R_{e}}{15}-R_{e}$

$h=\dfrac{R_{e}}{15}$

Hence, The maximum height is $\dfrac{R_{e}}{15}$

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Topic : escape velocity

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