Question

If a man weighs $60kg$ on the surface of the earth, the height above the surface of the earth where his weight is $30kg$ is

• A. $0.41R$
• B. $\frac{1}{2}R$
• C. $R/{\left(2\right)}^{\frac{1}{2}}$
• D. $R/2$

Answer 1

The correct option is A

$0.41R$

Step 1 Given,

Weight of object on the surface of earth = $60Kg$

Weight of object above a height h = $30Kg$

Step 2 : Finding h

We know that,

$W=mg$

At surface of earth , $g$= $\frac{GM}{R^2}$$=9.8m/s²$ - (i)

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

Dividing equation (i) and (ii)

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

= $\frac{1}{{\left(1+{\displaystyle \frac{h}{R}}\right)}^2}$

Where,

$h=$Height

$R=$Radius

Now,

$\frac{g\text{'}}{g}$ = $\frac{1}{{\left(1+{\displaystyle \frac{h}{R}}\right)}^2}$

$\frac{m{g}^{\text{'}}}{mg}=\frac{30}{60}=\frac{1}{{\left(1+{\displaystyle \frac{h}{R}}\right)}^2}$

$1+\frac{h}{R}=1.414$

$\frac{h}{R}$ $=1.414-1$

$\frac{h}{R}$ $=0.414$

$h=0.414R$

Hence, option (a) is correct.