# Question #d58fd

##### 1 Answer

May 14, 2016

#### Answer:

#### Explanation:

For small angle of deviation

#T=2pisqrt(L/g)# .......(1)

where#L# is the length of the pendulum and#g# is the local gravity.

- We know that seconds pendulum is a pendulum which has its period of two seconds. Its each swing takes one second,
#T=2s# - Also local acceleration due to gravity
#g=G (M_e)/R_e^2# .

#G# is constant#=6.67408 xx 10^-11 m^3 kg^-1 s^-2# .#M_e# is mass and#R_e# is radius of earth respectively.

Now the acceleration due gravity of the planet is

#g_p=G (M_p)/R_p^2#

Inserting given quantities we get

#g_p=G (2M_e)/(2R_e)^2#

#=>g_p=G (M_e)/(2(R_e)^2)# ,

in terms of#g#

#g_p=g/2#

Inserting in (1)

#T_p=2pisqrt(L/g_p)# ........(2)

in terms of

#T_p=2pisqrt(2L/g)#

Using (1)

#T_p=sqrt2T#

#=>T_p=2sqrt2s#