Question

An astronaut orbiting in a spaceship round the earth has a centripetal acceleration of $2.45m/s^2$. The height of spaceship from earth's surface is ($R=$radius of earth)

• A. $3R$
• B. $2R$
• C. $R$
• D. $R/2$

Answer 1

The correct option is C $R$

Given that,

Acceleration $a=2.45m/s^2$

Mass of earth $M=6\times {10}^{24}kg$

Radius of earth $R=6\times {10}^{6}m$

Gravitational constant $G=6.67\times {10}^{-11}N{m}^2/kg$

Now centripetal acceleration is

$a_c=\frac{v^2}{R}$

$a_c=\frac{{\left(\sqrt{\frac{GM}{R}}\right)}^2}{R}$

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

$a_c=\frac{GM}{{(R+h)}^2}$

Now, put the values

$2.45=\frac{6.67\times {10}^{-11}\times 6\times {10}^{24}}{{(6\times {10}^6+h)}^2}$

${(6\times {10}^6+h)}^2=\frac{6.67\times {10}^{-11}\times 6\times {10}^{24}}{2.45}$

${(6\times {10}^6+h)}^2=\frac{40.02\times {10}^{13}}{2.45}$

${(6\times {10}^6+h)}^2=1.6334\times {10}^{14}$

$6\times {10}^6+h=\sqrt{1.6334\times {10}^{14}}$

$h=1.278\times {10}^7-6\times {10}^6$

$h=(12.78-6)\times {10}^6$

$h=6.78\times {10}^6$

$h=6.8\times {10}^{6}m$

So, $h\cong R$

Hence, the height of spaceship is $R$