What is the magnitude of the gravitational force between the earth and a 1 kg object on its surface? (Mass of the earth is 6 × 1024 kg and radius of the earth is 6.4 × 106m.)

Asked by Vishal kumar | 1 year ago |  237

1 Answer

Solution :-

From Newton’s law of gravitation, we know that the force of attraction between the bodies is given by

F = \( \frac{(Gm_1m_2)}{r^2}\)

Here

m1 = mass of earth = 6.0 \( \times\)1024kg

m2 = mass of the body = 1kg

r = distance between the two bodies

Radius of Earth = 6.4 \( \times\) 106m

G = Universal gravitational constant = 6.67 \( \times\)10-11 Nm2kg-2

By substituting all the values in the equation

F =  \(\frac{(Gm_1m_2)}{r^2}\)

\(= \frac{6.67\times 10^{11}(6.0\times 10^{24}\times 1)}{(6.4\times 10^6)^2}\)

F = 9.8 N

This shows that Earth exerts a force o 9.8 N on a body of mass 1 Kg. The body will exerts an equal force of attraction of 9.8 N on the Earth.

Answered by Shivani Kumari | 1 year ago

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