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1. A spherical mass of 20 kg lying on the surface ofthe earth is attracted by another spherical mass of150 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»

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.


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