Question

Mass of earth is 6*10^24 and radius of earth 6.4*10^6 the magnitude of force between the mass of 1kg and the earth is

Answer 1

Here we are given,

Mass of earth (M) = 6 × 10²⁴ kg

mass of object (m) = 1 kg

Radius (r) = 6.4 × 10^6 m

Here we take the gravitational constant G as 6.67 × 10 -¹¹

Now, we know that,

F = GMm/r²

=> F =
$\frac{6.67 \times {10}^{ - 11} \times 6 \times {10}^{24} \times 1}{(6.4 \times {10}^{6} ) {}^{2} }$

$\frac{40.02 \times {10}^{13} }{( {6.4}^{2}) \times ( {10}^{6}) {}^{2} } \\ \\ = > \frac{40.02 \times {10}^{13} }{40.96 \times {10}^{12} } \\ \\ = > \frac{40.02}{40.96} \times 10$

=> 0.00977 × 10 (approx)

=> 0.0098 ×10 (approx)

=> 0.098 Newton Force

Answer 2

_/\_ Hello mate__ here is your answer -

Mass of Earth, M = 6 × 10^24 kg

Mass of object, m = 1 kg

Universal gravitational constant,

G = 6.7 × 10^−11 Nm^2 kg^−2

r = radius of the Earth (R) ( ob. is on the surface of earth)

r = R = 6.4 × 10^6 m

Therefore, the gravitational force

F = GMm / r^2

$=(6.7×10^-7×6×10^24 ×1)/(6.4×10^4)^2$

= 9.8 N

I hope, this will help you.☺

Thank you______❤

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