At what depth below the surface of the earth | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The acceleration due to gravity at height $h$ is given by $g_h = g \left( \frac{R}{R+h} \right)^2$. Given $h = 1600 \ km$ and $R = 6400 \ km$,we h

Vedclass · Accepted Answer

$4.8 	imes 10^6 \ m$

Vedclass · Answer

(A) The acceleration due to gravity at height $h$ is given by $g_h = g \left( \frac{R}{R+h} ight)^2$. Given $h = 1600 \ km$ and $R = 6400 \ km$,we have $g_h = g \left( \frac{6400}{6400+1600} ight)^2 = g \left( \frac{6400}{8000} ight)^2 = g \left( \frac{4}{5} ight)^2 = 0.64g$. The acceleration due to gravity at depth $d$ is given by $g_d = g \left( 1 - \frac{d}{R} ight)$. According to the problem,$g_d = \frac{1}{2} g_h = \frac{1}{2} (0.64g) = 0.32g$. Substituting this into the depth formula: $0.32g = g \left( 1 - \frac{d}{R} ight)$. $0.32 = 1 - \frac{d}{R} \Rightarrow \frac{d}{R} = 1 - 0.32 = 0.68$. $d = 0.68 	imes 6400 \ km = 4352 \ km = 4.352 	imes 10^6 \ m$. Note: If using the approximation $g_h \approx g(1 - 2h/R)$,then $g_h = g(1 - 3200/6400) = 0.5g$. Then $g_d = 0.5 g_h = 0.25g$. $0.25g = g(1 - d/R) \Rightarrow d/R = 0.75 \Rightarrow d = 0.75 	imes 6400 \ km = 4800 \ km = 4.8 	imes 10^6 \ m$. Given the options,the approximation method is intended.

At what depth below the surface of the earth will the acceleration due to gravity be half of its value at $1600 \ km$ above the surface of the earth?

Similar Questions

Acceleration due to gravity on the surface of Earth is $g$. If the diameter of Earth is reduced to one-third of its original value and mass remains unchanged,then the acceleration due to gravity on the surface of the Earth is . . . . . . $g$.

For a sphere of mass $M$ and radius $R$,if the gravitational forces at distances $r_1$ and $r_2$ from the center are $F_1$ and $F_2$ respectively,then:

The mass and diameter of a planet have twice the value of the corresponding parameters of Earth. Acceleration due to gravity on the surface of the planet is ........ $m/s^2$.

The value of gravitational acceleration $g'$ at a height $h$ above the earth's surface is $\frac{g}{4}$. Then,what is the value of $h$ in terms of the earth's radius $R$?

If the mass of the Sun were ten times smaller and the universal gravitational constant were ten times larger in magnitude,which of the following is not correct?

Vedclass Test Series

Exam Paper Generator

Online Exam Module