Question

Two masses, m1 and m2 are separated by a distance, r. The force of attraction between the two masses is F. A) If m1 increases by 8, how does F change? B) If r is halved, m1 increases by 4 how would F change? C) If r is not changed but both masses increase by a factor 8, how would F change? D) If m1 and m2 remain unchanged, and r changes to 3/5 of the original distance, how would F change?

Answer 1

Answer:

body with an initial velocity x moves with an uniform acc y. Plot its velocity time graph

Answer 2

(A.) Therefore If m₁ increases by 8, then F becomes '8F'.

(B.) Therefore if d is halved, m₁ increases by 4 then F becomes '8F'.

(C.) Therefore if d is not changed but both masses increase by a factor of 8, then F becomes '16F'.

(D.) Therefore if m₁ and m₂ remain unchanged, and d changes to 3/5 of the original distance, then F becomes '25F/9'.

Given:

The masses = m₁, and m₂

The distance between 2 masses = d

The force of attraction between the 2 masses = F

To Find:

(A.) If m1 increases by 8, then F becomes?

(B.) If r is halved, m1 increases by 4 then F becomes?

(C.) If r is not changed but both masses increase by a factor of 8, then F becomes?

(D.)  If m1 and m2 remain unchanged, and r changes to 3/5 of the original distance, then F becomes?

Solution:

The given question can be solved very easily as shown below.

The Force of attraction F = Gm₁m₂/d²

Where G = Gravitational Constant

So F ∝ m₁m₂/d²          (i.)

(A.) If m1 increases by 8, then F becomes:

⇒ F¹ ∝ m₁¹m₂¹/(d²)¹

Given m₁¹ = 8m₁, m₂¹ = m₂, and d¹ = d

Then F¹ ∝ 8m₁m₂/d²          (ii.)

⇒ equation (ii.)/equation (i.) ⇒ F¹/F = ( 8m₁m₂/d² )/( m₁m₂/d² )

⇒ F¹/F = 8 ⇒ F¹ = 8F

Therefore if m₁ increases by 8, then F becomes '8F'.

(B.) If d is halved, m1 increases by 4 then F becomes:

⇒ F¹ ∝ m₁¹m₂¹/(d²)¹

Given m₁¹ = 4m₁, m₂¹ = m₂, and d¹ = d/2

Then F¹ ∝ 4m₁m₂/d²/4 ⇒ F¹ ∝ 8m₁m₂/d²          (iii.)

⇒ equation (iii.)/equation (i.) ⇒ F¹/F = ( 8m₁m₂/d² )/( m₁m₂/d² )

⇒ F¹/F = 8 ⇒ F¹ = 8F

Therefore if d is halved, m₁ increases by 4 then F becomes '8F'.

(C.) If r is not changed but both masses increase by a factor of 8, then F becomes:

⇒ F¹ ∝ m₁¹m₂¹/(d²)¹

Given m₁¹ = 8m₁, m₂¹ = 8m₂, and d¹ = d

Then F¹ ∝ 16m₁m₂/d²          (iv.)

⇒ equation (iv.)/equation (i.) ⇒ F¹/F = ( 16m₁m₂/d² )/( m₁m₂/d² )

⇒ F¹/F = 16 ⇒ F¹ = 16F

Therefore if r is not changed but both masses increase by a factor of 8, then F becomes '16F'

(D.)  If m1 and m2 remain unchanged, and r changes to 3/5 of the original distance, then F becomes:

⇒ F¹ ∝ m₁¹m₂¹/(d²)¹

Given m₁¹ = m₁, m₂¹ = m₂, and d¹ = 3d/5

Then F¹ ∝ m₁m₂/9d²/25 ⇒  F¹ ∝ 25m₁m₂/9d²          (v.)

⇒ equation (v.)/equation (i.) ⇒ F¹/F = ( 25m₁m₂/9d² )/( m₁m₂/d² )

⇒ F¹/F = 16 ⇒ F¹ = 25F/9

Therefore if m₁ and m₂ remain unchanged, and d changes to 3/5 of the original distance, then F becomes '25F/9'.

(A.) Therefore If m₁ increases by 8, then F becomes '8F'.

(B.) Therefore if d is halved, m₁ increases by 4 then F becomes '8F'.

(C.) Therefore if d is not changed but both masses increase by a factor of 8, then F becomes '16F'.

(D.) Therefore if m₁ and m₂ remain unchanged, and d changes to 3/5 of the original distance, then F becomes '25F/9'.

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