Question

If the earth is squesed gravitationally to its half of its presnt radiuus the duration of the day will be

Answer 1

Answer:

You can solve it using Kepler's Third law, according to which,

(T2/T1)^2 = (R2/R1)^3 (eq.1)

Where, T2 = time period of Earth of its radius being reduced,

and R2 = the final radius

Also, T1 and R1 are the original time period and radius respectively.

Now, it's given that radius is reduced to half of its original value,

So,

R2 = R1/2

And T1 = 365 days = 365 × 24 h = 8760 h

Now, putting the above values in equation 1,

(T2/8760)^2 = (R1/2/R1)^3

T2^2 = (1/2)^3 × (8760)^2

T2 = Root 8760^2/8

T2 = 8760/2 × root 2

T2 =4380/root2 = 4380/1.414

(since root 2 = 1.414)

T2 = 3097.5 h

Dividing by 24 h,

T2 = 3097.5 / 24 = 129.04 days.

Hence the new time period will be approximately 129 days.