Question #0944f

Jan 24, 2018

$2.45 \frac{m}{s} ^ 2$

Explanation:

If the value at the earth's surface is g, we knew from the law of universal gravitation that this can be calculated from the mass of the earth (M), the gravitational constant (G), and the radius of the earth (R).

$g = G \frac{M}{R} ^ 2$

At a radius of 2*R the new value for acceleration (let's call it a) will be equal to:

$a = G \frac{M}{2 R} ^ 2$
$a = G \frac{M}{4 {R}^{2}}$

With a little algebra we can see that this can be written as:
$a = G \frac{M}{{R}^{2}} \cdot \frac{1}{4} = g \cdot \frac{1}{4} = \frac{g}{4}$

Solving for a using the value we know for g.
$a = \frac{9.8}{4} = 2.45 \frac{m}{s} ^ 2$

Jan 24, 2018

Law of Universal Gravitation states that the force of attraction ${F}_{G}$ between two bodies of masses ${M}_{1} \mathmr{and} {M}_{2}$ is directly proportional to the product of masses of the two bodies. It is also inversely proportional to the square of the distance $r$ between the two.

${F}_{G} = G \frac{{M}_{1.} {M}_{2}}{r} ^ 2$ .......(1)
where $G$ is the Gravitational constant.
It has the value $6.67408 \times {10}^{-} 11 {m}^{3} k {g}^{-} 1 {s}^{-} 2$

If one of the objects is earth, mass ${M}_{e}$, force between an object of mass $m$ located at the earth's surface is

${F}_{G} = G \frac{{M}_{e} . m}{R} ^ 2$ ......(2)
where $R$ is radius of earth.

Comparing with equation

$\text{Weight} = m g$, we get
$g = \frac{G {M}_{e}}{R} ^ 2 = 9.8 m {s}^{-} 2$ .....(3)

When the object is located at a height equal to radius of earth, its distance from centre of the earth is $R + R = 2 R$. (2) becomes

${F}_{G} = G \frac{{M}_{e} . m}{2 R} ^ 2$

If ${g}^{'}$ is acceleration due to gravity at this location we get

${g}^{'} = \frac{1}{4} \frac{G {M}_{e}}{R} ^ 2$ .........(4)

Dividing (4) by (3) and rearranging we get

${g}^{'} = \frac{g}{4}$
${g}^{'} = \frac{9.8}{4} = 2.45 m {s}^{-} 2$