Question

What is the value of acceleration due to gravity at a height equal to half the radius of earth, from the surface of earth? [take $g = 10\,m{s^{ - 2}}$ on earth’s surface]

Answer 1

Let us know about the acceleration.The net acceleration imparted to objects due to the combined effects of gravitation (from mass distribution within Earth) and centrifugal force (from the Earth's rotation) is symbolised by $g$.

Complete step by step answer:
Gravity acceleration is a vector quantity with both magnitude and direction. Gravity would point directly towards the sphere's centre in a spherically symmetric Earth. Because the Earth's shape is significantly flatter, considerable variances in gravity direction exist:essentially the difference between geodetic and geocentric latitude. Local mass anomalies, such as mountains, create smaller deviations, known as vertical deflection.

Let $R$ be the radius of the earth from its surface, and $h$ be the height equivalent to half the radius of the earth from its surface and $M$ is the mass of the earth.The acceleration due to gravity on the earth's surface is given as,
$g = \dfrac{{GM}}{{{R^2}}}$
The acceleration due to gravity at a height is given as,
${g^1} = \dfrac{{GM}}{{{R^2}}}$
$\Rightarrow {g^1} = \dfrac{{GM}}{{{{\left( {\dfrac{R}{2}} \right)}^2}}}$
$\Rightarrow {g^1} = 4g$
$\Rightarrow {g^1} = 4 \times 10$
$\therefore {g^1} = 40\,m{s^{ - 2}}$

Hence, the value of gravity-induced acceleration at a height is $40\,m{s^{ - 2}}$.

Note: This acceleration is expressed in SI units as metres per second squared (in symbols,$m{s^{ - 2}}$) or newtons per kilogramme $\left( {N/kg{\text{ }}or{\text{ }}Nk{g^{ - 1}}} \right).$ Gravitational acceleration near the Earth's surface is around $9.81{\text{ }}m{s^{ - 2}}$, which means that, excluding the effects of air resistance, the speed of an object falling freely will increase by roughly $9.81{\text{ }}m{s^{ - 1}}$ per second.