SATHEE: Gravitation Question 66

Gravitation Question 66

Question:At what depth below the surface of the earth, acceleration due to gravity g will be half itsvalue 1600 km above the surface of the earth[Pb. PMT 2004]

Options:

A) $[4.2 \times 10^{6} m]$

B) $[3.19 \times 10^{6} m]$

C) $[1.59 \times 10^{6} m]$

D) None of these

Show Answer

Answer:

Correct Answer: A

Solution:

Radius of earth R = 6400 km \ $[h = \frac{R}{4}]$

Acceleration due to gravity at a height h

$[g_{h} = g {(\frac{R}{R + h})}^{2}]$

$[= g {(\frac{R}{R + \frac{R}{4}})}^{2}]$

$[= \frac{16}{25} g]$ At depth ’d’ value of acceleration due to gravity $[g_{d} = \frac{1}{2} g_{h}]$

(According to problem) Þ $[g_{d} = \frac{1}{2} (\frac{16}{25}) g]$

$[\Rightarrow g (1 - \frac{d}{R})]$

$[= \frac{1}{2} (\frac{16}{25}) g]$

By solving we get $[d = 4.3 \times 10^{6} m]$

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Gravitation Question 66

Question:At what depth below the surface of the earth, acceleration due to gravity g will be half itsvalue 1600 km above the surface of the earth[Pb. PMT 2004]

Options:

Answer:

Solution:

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