You are an astronaut on a planet of mass M=3.3 M_earth and equatorial radius R=2

Question

You are an astronaut on a planet of mass M=3.3 M_earth and equatorial radius R=2.1 R_earth, in Earth units. You set up a pendulum of length L_0=54.2cm on the surface of that planet. The planet rotates in a time T=17 hrs. What is the acceleration a_gray of gravity at the planet's surface? What is the period P of the pendulum? How many (complete) oscillations n of the pendulum are there in the planet's day? By how much, Delta a, is the acceleration of gravity reduced at the equator due to the rotation of the planet?

Explanation / Answer

Solution: Mass of the Earth Mearth = 5.98*1024 kg

Radius of Earth Rearth = 6.371*106 m

Mass of the planet M = 3.3* Mearth = 3.3*5.98*1024 kg = 1.9734*1025 kg

Radius of the planet R = 2.1* Rearth = 2.1*6.371*106 m = 1.33791*107 m

Length of the pendulum L = 54.2 cm = 0.542 m

Length of the day of the planet T =17 hrs = (17hrs)*(60min/1hrs)(60sec/1min) = 61200 seconds.

Part (a)

acceleration due to gravity at the planet surface is given by,

agrav = G*M/R2

agrav = (6.67*10-11 m3/kg.s2)*( 1.9734*1025 kg)/( 1.33791*107m)2

agrav = 7.3534 m/s2

Thus the acceleration agrav of gravity at the planet surface is 7.35 m/s2.

agrav = 7.35 m/s2

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Part (b)

Period of the pendulum is given by,

P = 2**[L/ agrav]

P = 2*3.141*[(0.542m)/(7.3534m/s2)]

P = 1.7058 sec.

Hence the period of the pendulum is 1.706 seconds.

P = 1.706 sec

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Part (c) Since the one day of the planet has 17 hrs that is 61200 seconds and since the period of the pendulum is 1.7058 seconds, the number of oscillations n is given by,

n = T/P

n = 61200sec / 1.7058sec

n = 35877.59

Hence the pendulum completes 35877 oscillations in the planet’s day.

n = 35877

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Part (d)

Since the planet rotates with an angular velocity = 2/T, there exists a centripetal acceleration 2R due rotation which acts in the opposite direction of the acceleration of gravity. Thus change in the acceleration a is given by,

a = 2R

a = (2/T)2R

a = (2*3.141rad/61200s)2*(1.33791*107 m)

a = 0.141 m/s2

Thus the at the equator, the gravity is reduced by a = 0.141 m/s2

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