Question

1. A spherical mass of 20 kg lying on the surface of
the earth is attracted by another spherical mass of
150 kg with a force equal to the weight of 0.25 mg.
The centres of the two masses are 30 cm apart.
Calculate the mass of the earth. Radius of the earth
= 6 x 10^6m.
​

Answer 1

Answer:

$5.289 \times {10}^{24} kg$

Explanation:

Given:

$m_{1} = 20 \: kg \\ \\ m_2 = 150 \: kg$

F = 0.25 mg

$= 0.25 \times {10}^{ - 6} \\ (1mg \: = {10}^{ - 6} kg)$

r = 30 cm

F = m×g

$= \frac{30}{100} \: m \\ \\ = {10}^{ - 1} m$

We know that,

$F \: = \: \frac{G. m_1 .m_2 }{ {r}^{2} } \\ \\ G = \frac{f {r}^{2} }{ m_1 . m_2 } \\ \\ = \frac{0.25 \times {10}^{ - 6} \times 9.8 \times {10}^{ - 1} \times {10}^{ - 1} }{20 \times 150} \\ \\ G = 6.67 \times {10}^{ - 11} N. {m}^{2} .kg {}^{ - 2}$

Now , by using the formula of g (acceleration due to gravity)

$g \: = \frac{GM}{ {R}^{2} }$

Where G = Gravitational constant,

M = mass of the earth

R = radius of the earth.

Here , we know the value of "g'' on the earth, i.e.9.8 m/s ².

$9.8 = \frac{6.67 \times {10}^{ - 11} \times M }{(6 \times {10}^{6}) {}^{2} } \\ \\ 9.8 = \frac{6.67 \times {10 }^{ - 11} \times M}{36 \times {10}^{12} } \\ \\ 9.8 \times 36 \times {10}^{12} \: = 6.67 \times {10}^{ - 11} \times M \\ \\ \frac{9.8 \times 36 \times {10}^{12} }{6.67 \times {10}^{ - 11} } = M \\ \\ M \: = 5.289 \times {10}^{24} kg$.

Answer 2

m1=20Kg

m2=150kg

0.25mg=o.25x10^-6

r=0.30m

F=mxg