Question

An experiment using the Cavendish balance to measure the gravitational constant G found that mass of $0.800\,kg$ attracts another sphere of mass $4 \times {10^{ - 3}}\,kg$ with a force of $1.30 \times {10^{ - 10}}\,N$ when the distance between the centers of the spheres is $0.0400\,m.$ The acceleration due to gravity at the earth’s surface is $9.80\,m{\sec ^{ - 2}}$ and the radius of earth is $6380\,km$ The mass of the earth from this data is:-
A. $8 \times {10^{24}}kg$
B. $8 \times {10^{23}}kg$
C. $6 \times {10^{23}}kg$
D. $6 \times {10^{24}}kg$

Answer 1

Hint: In order to solve this question we need to understand Newton's law of gravitation. So according to Newton, gravity is a force with which it attracts mass above earth and this force is called gravitational force. This force is directly proportional to the product of two masses and inversely proportional to the square of distance between them. But according to Einstein, gravity is not a force rather it is a consequence of the space time curve.

Complete step by step answer:
Let the mass of the object be, $m$. Also let the mass of the sphere be, ${M_1}$ having radius ${R_1}$. Also let the mass of earth be, ${M_e}$ having radius ${R_e}$.
According to question, $m = 0.8kg$
And, ${M_1} = 4 \times {10^{ - 3}}kg$
Also, ${R_1} = 0.04m$ and ${R_e} = 6380 \times {10^3}m$
Let F be the force of attraction between mass of object and Mass of sphere is given in accordance with Newton’s law as,
$F = \dfrac{{GmM}}{{{R_1}^2}}$
Since, $F = 1.30 \times {10^{ - 10}}N$ (given force)
Putting values we get,
$1.30 \times {10^{ - 10}}N = \dfrac{{G(0.8kg)(4 \times {{10}^{ - 3}}kg)}}{{{{(0.04m)}^2}}}$
$\Rightarrow 1.30 \times {10^{ - 10}}N = G(2k{g^2}{m^{ - 2}})$
So value of G is,
$G = \dfrac{{1.30 \times {{10}^{ - 10}}N}}{{2k{g^2}{m^{ - 2}}}}$
$\Rightarrow G = 0.65 \times {10^{ - 10}}k{g^{ - 1}}{m^3}{\sec ^{ - 2}}$

Now, if the object attracts by the earth, we know force with which it attracts is given as,
$f = mg$
Where, “g” is acceleration due to gravity and, $g = 9.8m{\sec ^{ - 2}}$.
Also from Newton’s law, the value of gravitational force is given as,
$f = \dfrac{{Gm{M_e}}}{{{R_e}^2}}$
Putting value of “f”, “m” and ${R_e}$ we get,
$mg = \dfrac{{m(0.65 \times {{10}^{ - 10}}){M_e}}}{{{{(6380 \times {{10}^3})}^2}}}$
So Mass of earth is,
${M_e} = \dfrac{{g{{(6380 \times {{10}^3})}^2}}}{{0.65 \times {{10}^{ - 10}}}}$
$\Rightarrow {M_e} = (9.8) \times 6262.21 \times {10^{20}}$
$\Rightarrow {M_e} = 61369.7 \times {10^{20}}kg$
$\therefore {M_e} = 6.13 \times {10^{24}}kg$
So the mass of earth is, ${M_e} = 6.13 \times {10^{24}}kg$.

So the correct option is D.

Note: It should be remembered that the Cavendish device is used to measure Gravitational constant or Gravitational proportionality constant “G”. Using this method we can calculate the mass of any planet, comet or any other celestial body. Newton’s laws are only valid in the inertial frame of reference but Einstein’s law of Special theory of relativity valid in all frames of reference.