Question

what would be the average duration of year if the distance between the sun and the earth becomes( a) thrice the present distance (b) twice the present distance​

Answer 1

Answer:

T  

2

αR  

3

 

If r get double⇒R  

2

​  

=2R  

1

​  

 

T  

1

​  

 

T  

2

​  

 

​  

=(  

R  

1

​  

 

R  

2

​  

 

​  

)  

2

3

​  

 

=2  

2

3

​  

 

 

⇒T  

2

​  

=(2  

2

​  

)T  

1

​  

=(2  

2

​  

)365days=1032days

Answer 2

Given:

• $T_1$ = 365 days

To Find:

• Duration of the year when the

(a) Distance between the sun and earth is thrice the present distance.

(b) Distance between the sun and the earth is twice the present distance.

Solution:

(a)

• Let $r_1$ = present distance between the sun and the earth.
• $r_2$ = 3$r_1$
• To find the duration of the year we are using Kepler's law of period.
• $T_1^{2} \alpha r_1^{3}$ and $T_2^{2} \alpha r_2^{3}$
• The above formula can be written as $\frac{T_1^{2} }{T_2^{2} } = \frac{r_2^{3} }{r_1^{3} }$ = $\frac{T_2}{T_1}=(\frac{r_2}{r_1}) ^{3/2}$  = $(\frac{3r_1}{r_1} )^{3/2}$
• $\frac{T_2}{T_1}=3^{3/2} = \sqrt{27}$
• $T_2 = T_1*\sqrt{27} = 365*(3\sqrt{3}) = 365*5.196 = 1896.5 days.$

(b)

• Let $r_1$ = present distance between the sun and the earth.

• $r_2 = 2r_1$

• To find the duration of the year we are using Kepler's law of period.

• $T_1^{2} \alpha r_1^{3}$ and $T_2^{2} \alpha r_2^{3}$  
• The above formula can be written as $\frac{T_1^{2} }{T_2^{2} } = \frac{r_2^{3} }{r_1^{3} }$ = $\frac{T_2}{T_1}=(\frac{r_2}{r_1}) ^{3/2}$  = $(\frac{2r_1}{r_1} )^{3/2}$
• $\frac{T_2}{T_1}=2^{3/2} = \sqrt{8} = 2\sqrt{2}$
• $T_2 = T_1*2\sqrt{2}$
• $T_2 = 365*2\sqrt{2} = 365*2(1.41) = 356*2.82 = 1029.3 days.$

(a) $T_2 = 1896.5 days$

(b) $T_2 = 1029.3days$