Question

The acceleration due to gravity at a height $\text{'}h\text{'}$ from the surface of the earth is 96% less than its value on the surface then $h$= ………… R, where R is the radius of the earth.

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is D

$4$

Step 1: Given data

The radius of the earth is R.

Step 2: Calculating height h.

The acceleration due to gravity at a height h above the earth's surface can be calculated as:$g_h=\frac{GM}{{(R+h)}^2}$

Here, M is the mass of the earth.

R is the radius of the earth.

G is the universal Gravitational constant.

Now, at the surface,

$g=\frac{GM}{R^2}$

Now if we rearrange $g_h$ using the above equations as follows

$g_h=\frac{GM}{R^2}\times \frac{R^2}{{\left(R+h\right)}^2}$

$g_h=g{\left(\frac{R}{R+h}\right)}^2$

According to the question $g_h$ is 96% less than its value on the surface

i.e. $100-96=4\%$ of its value on surface.

Thus, $g_h=0.04g$

$g{\left(\frac{R}{R+h}\right)}^2=0.04g$

$\frac{R}{R+h}=\sqrt{0.04}=0.2$

$\frac{R}{R+h}=\frac{1}{5}$

$5R=R+h$

$h=5R-R$

Step 3: Final answer

$h=4R$

So, the acceleration due to gravity is 96% less than its value on the surface of the earth at a height of 4R, where R is the radius of the earth.

Thus the correct answer is (d).