Question

The value of $g$ on the earth's surface is $980cm/se{c}^2$ . Its value at a height of $64km$ from the earth's surface in $cm{s}^{-2}$ is

• A. $960.40$
• B. $984.90$
• C. $982.45$
• D. $977.55$

Answer 1

The correct option is A

$960.40$

Explanation for the correct answer:-

Step 1. Given Data,

$g=980cm/s^2$

$h=64km$

Step 2. Formula used

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

$g\text{'}=\frac{GM}{{(R+h)}^2}$

$g$ is the acceleration due to gravity at the earth's surface, $G$ is the gravitation constant, $M$ is the mass of the body, $R$ is the radius.

$g\text{'}$ is the acceleration due to gravity at a height $64km$ from the earth's surface.

$h$is the height from the earth's surface.

Step 3: Calculating the Value of g at the earth's surface,

Given, $g=980cm/s^2$

Converting $g$ from $cm$ to $m$ by dividing $980$ by $100$, as $1m=100cm$

At the earth's surface, $g=\frac{GM}{R^2}=9.8m/s^2$

$\frac{GM}{{(6400\times {10}^3)}^2}=9.8m/s^2......1$

Step 4: Evaluating $g\text{'}$ at a height $64km$ from the earth's surface,

$g\text{'}$ is the gravity taken at a height $64kms$ from earth's surface,

$\frac{GM}{{\left(\right(6400+64)\times {10}^3)}^2}=g\text{'}.......\left(2\right)$

On dividing $2$ by $1$, we get

${\left(\frac{6400}{6464}\right)}^2=\frac{9.8}{g\text{'}}$

$g\text{'}=9.8\times {\left(\frac{6464}{6400}\right)}^2$

$$\begin{gathered} g\text{'}=9.6040m/s^2 \\ g\text{'}=960.40cm/s^2 \end{gathered}$$

Hence option A is the correct option.