Question

3. 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 x 10²⁴ kg and radius of the earth is 6.4 x 10⁶m.)
​

Answer 1

Answer:

use formula f=Gm1m2/r^2

G=6.67×10^-11

Explanation:

putting value

f=(6.67×10^-11 × 1 × 6×10^24) / (6.4×10^6)^2

by using scientific calculator

f=9.77N is answer

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Answer 2

Answer :

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.