Question

What is the magnitude of the gravitational force between the
earth and a 3 kg object on its surface? (Mass of the earth is
6 x 10²⁴ kg and radius of the earth is 6.4 x 10⁶m.)

please solve step by step.. please..... ​

Answer 1

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There will be a gravitational force of approximately 9.77 Newton b/w earth and 1 kg object on earth's surface.

Explanation :

Given,

Mass of Earth, M = 6 × 10²⁴ kg

Radius of Earth, R = 6.4 × 10⁶ m

mass of object on the earth's surface, m = 1 kg

To find,

Gravitational force between Earth and 1 kg object on Earth's Surface , F = ?

so,

Using formula for Gravitational force

F = G M m / R²

[ where F is the gravitational force b/w two objects of masses M and m , G is the universal gravitational constant ] and R is the distance b/w those two objects ]

Taking , G = 6.67 x 10⁻¹¹ m³ kg⁻¹ s⁻²

Using formula for Gravitational force

→ F = G M m / R²

→ F = (6.67 × 10⁻¹¹) (6 × 10²⁴) (1) / (6.4 × 10⁶)²

→ F = [ 6.67 × 6 × 10¹³ ] / ( 6.4 × 6.4 × 10¹² )

→ F = ( 40.02 × 10¹³ ) / ( 40.96 × 10¹² )

→ F =  0.977 × 10

→ F ≈ 9.77 N

therefore,

There will be a gravitational force of approximately 9.77 Newton b/w earth and 1 kg object on earth's surface.

Answer 2

Answer:

Given : Mass of earth M=6×10^24 kg

Mass of object m=1 kg

Radius of earth R=6.4×10^6m

Force of gravitation between them, F= R ^2/ GMm

F= (6.4×10 ^6 ) ^2/

6.67×10 ^−11 ×(6×10 24 )×1 =9.77 N