Question

A man who weigh 75 kg is on the surface of the earth whose mass is 6*10^24 kg. If the radius of the earth is 6380 km, calculate the force of attraction between them.

Answer 1

hi dear

your answer is

Given : Mass of the earth (m1)=6×1024kg

Radius of the earth (R ) =6.4×106m ,

Mahendra's mass (m2)=75kg

Gravitational constant (G) =6.67×10-11Nm2/kg2

To find : Gravitational force (F)

F=Gm1m2R2

F=6.67×10-11×75×6×1024(6.4×106)2

=6.67×75×66.4×6.4

=733N

hopes its helps you

Answer 2

Answer:

The force of attraction between the man and the earth is found to be 737.72 Newtons.

Explanation:

Newton’s Law of Gravitation states that each particle attracts every other particle in the cosmos with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance present between them.

This statement gives us the expression as:

$F=-\frac{Gm_1m_2}{r^2}$

Now, we are given the masses of two bodies considered as 75 kg and $6\times10^{24}$ kg respectively. The radius of the earth is the distance present between the earth's core and the man present at its surface is the radius of the earth which is given as 6380 km.

Also, we know that the Gravitational constant, 'G' has the value as:

$6.673 \times 10^{-11} N m^2/kg^2.$

Substituting all these values, we get:

$F=-\frac{6.673\times 10^{-11}\times75\times 6\times10^{24}}{(6380\times1000)^2}$

$F=-\frac{3002.85\times 10^{13}}{407044\times10^8}$

Simplifying it, we get:

$F=-0.0073772122\times10^5$

or we can say:

$F=-737.72\ N$

Thus, the force of attraction between them is found to be 737.72 Newtons.

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