In this problem, be especially wary of round-off; keep 3 sigfigs in each answer

Question

In this problem, be especially wary of round-off; keep 3 sigfigs in each answer and throughout your calculations.

Based on the following data about planet X (which orbits around the Sun):

Planet X's distance from Sun = 5*1012 m
Planet X's radius = 1.6*106 m
Planet X's mass = 8.4*1022 kg

a.) Find gx, the size of the acceleration due to gravity on the surface of Planet X.

b.) What is the weight of a 10 kg mass on the surface of Planet X?
(How does this compare to its weight on Earth?)

c.) How long would it take for a ball dropped from a height of 18 m to hit the ground?
(How does this compare to the time it would take on Earth?)

d.) At 3 of Planet X's radii above the planet's surface, what is gx?

e.) Find the orbital speed of Planet X around the Sun.

f.) How long is a year on Planet X? Express your answers in both seconds and Earth years:
seconds?Earth years?

Explanation / Answer

Given that distance h=5 x 1012 m radius R=1.6 x 106 m mass M=8.4 x 1022 kg a)the accleration due to gravity is gx=GM/R2                                                           =6.67 x 10-11 x 8.4 x 1022 /(1.6 x 106 )2                                                           =6.67 x 10-11 x 8.4 x 1022 /(1.6 x 106 )2                                                           =2.188 m/s2 which is 0.223 times the accleration due to gravity of earth b)mass m=10kg weight W=mgx which is 0.223 times the accleration due to gravity of earth b)mass m=10kg weight W=mgx                  =21.88N which is 0.223 times the weight on the earth c) given that height s=18m using kinematic relation s=ut+1/2at2                                        s=ut+1/2gxt2                                       18=1/2(2.188)t2                                      t=4.056sec where as on earth  s=ut+1/2gt2                              18=1/2(9.8)t2                              t=3.673sec So it take longer time for the objectto reach on planet earth than compared to that of earth d) at a height 3R the accleration due to gravity is gx =GM/(R+3R)2      =GM/16R2      =0.136m/s2 e) orbital speed vo=GMs/(Rs+h)    where mass of the sumn is Ms =1.99 x 1030 kg radius of the sun is Rs=6.96 x 108m                   vo =GMs/(Rs+h)                       =(6.67 x 10-11x1.99 x 1030 /(6.96 x 108+5 x 1012 )                     =5.151km/s f)period of reveloution t=2(Rs+h )/ vo     =2(6.96 x 108+5 x 1012 )/5.151 x 103    =6.099 x 109s     interms of earth years it take194.23 years    =6.099 x 109s     interms of earth years it take194.23 years      

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