3. What is the magnitude of the gravitational force between theearth a

1.	3. What is the magnitude of the gravitational force between theearth and a 1 kg object on its surface? (Mass of the earth is6 x 10²⁴ kg and radius of the earth is 6.4 x 10⁶m.)
Answer» 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.

3. What is the magnitude of the gravitational force between theearth and a 1 kg object on its surface? (Mass of the earth is6 x 10²⁴ kg and radius of the earth is 6.4 x 10⁶m.)

Answer»

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.