The height above surface of earth where the value of gravitational acceleration is one fourth of that at surface, will be

Question

The height above surface of earth where the value of gravitational acceleration is one fourth of that at surface, will be (A) (Re/4) (B) (Re/2) (C) (3 Re/4) (D) Re

Tardigrade Admin · Accepted Answer

gâ²=(g R2/(R + h)2) (g/4)=(g R2/(R + h)2) (1/4)=(R2/(R + h)2) (R + h)2=4R2 (R + h)=2R h=R=Re

Q. The height above surface of earth where the value of gravitational acceleration is one fourth of that at surface, will be

$\frac{R _{e}}{4}$

$\frac{R _{e}}{2}$

$\frac{3 R _{e}}{4}$

$R_{e}$

$g' = \frac{g R ^{2}}{( R + h ) ^{2}}$
$\frac{g}{4} = \frac{g R ^{2}}{( R + h ) ^{2}}$
$\frac{1}{4} = \frac{R ^{2}}{( R + h ) ^{2}}$
$(R + h)^{2} = 4 R^{2}$
$(R + h) = 2 R$
$h = R = R_{e}$

Q. The height above surface of earth where the value of gravitational acceleration is one fourth of that at surface, will be

4Re​​

2Re​​

43Re​​

Re​

g′=(R+h)2gR2​ 4g​=(R+h)2gR2​ 41​=(R+h)2R2​ (R+h)2=4R2 (R+h)=2R h=R=Re​

$\frac{R _{e}}{4}$

$\frac{R _{e}}{2}$

$\frac{3 R _{e}}{4}$

$R_{e}$

$g' = \frac{g R ^{2}}{( R + h ) ^{2}}$
$\frac{g}{4} = \frac{g R ^{2}}{( R + h ) ^{2}}$
$\frac{1}{4} = \frac{R ^{2}}{( R + h ) ^{2}}$
$(R + h)^{2} = 4 R^{2}$
$(R + h) = 2 R$
$h = R = R_{e}$