Planets in orbit have two kinds of rotational motions. The first is that the pla

ID: 3162187 • Letter: P

Question

Planets in orbit have two kinds of rotational motions. The first is that the planet rotates about an axis that (roughly) passes through its center of mass. The second is that the planet moves in an elliptical orbit around a star. For most of the planets in our solar system the eccentricity, or ellipticity of the orbit is quite small and it is not a bad approximation to think of the orbit as a circle with the Sun at the center. It should be obvious, or at least unsurprising that the radius of the orbit is much, much greater than the radius of the planet. In most, but not all cases, the period of the orbital motion (the yearly motion) is much longer than the rotational motion (the daily motion). We can think of the planet as having two components of angular momentum, one due to the daily motion, and one due to the yearly motion. Which component of angular momentum (as measured with respect to the axis of rotation that makes sense in each case) do you think is greater for a typical planet (i.e. if the year is much longer than the day)? Why do you think this would be the case? (No penalty for wrong answers here, if you provide an honest account of what you think and why). Look up any information you need to answer the following questions and clearly cite your sources such that the grader could easily verify where you found the information and would likely agree that the source is trustworthy. What is the rotational inertia of the Earth about an axis through its center of mass? What is the rotational inertia of the Earth through an axis that passes through the center of its orbit (assume the orbit is very nearly circular). When calculating the second quantity you have a choice. You can treat the earth as a sphere with a particular radius, or as a point particle. Justify your choice. Repeat part (b) for Jupiter, Mercury and Pluto. How much angular momentum does the Earth have about an axis through its center of mass as a result of its daily rotation? How much angular momentum does the Earth have about an axis through the center of its nearly circular orbit as a result of its yearly orbital motion? How do these quantities compare? Repeat (d) for Jupiter, Mercury and Pluto. Based on your calculations, what factor dominates the singular momentum calculation?

Explanation / Answer

a)For rotation of planet about own axis we will take as case 1 and revolution of planet about the star will take as case 2

case1) L1=I *w=2/5 * M *R2 *angular velocity of planet about own axis(w1)

{moment of inertia of sphere is 2/5 M R2}

case 2) L2= I *w= M*d2*angular velocity of planet about star(w2)

d>>R

w1>>w2

its a spin off between these two values

generally, for farther planets d>>R is a bigger deciding factor than w1>>w2 thus

L2>L1

b) rotational inertia of earth about own axis= I1=2/5 * M *R2

For earth, Radius: 6,371 km

Mass: 5.972 × 10^24 kg

source: Wikipedia

I1=9.69*1037kgm2

rotational inertia of earth about centre of its orbit around sun= I2=M*d2

d=149.60*106km

I2=1.336*1047kgm2

as d>>R, thus even if we take it as a sphere the change in quantity will me very minute. Thus its more practical to consider it as a point mass.

For Jupiter

For mercury

For pluto

69,911 km

2,440 km

1,187 km

778*10^6 km

70000km

5.90*10^9km

1.898 × 10^27 kg

3.285 × 10^23 kg

1.30900 × 1022 kg

3.71E+42

7.82303E+35

7.37736E+33

1.15E+51

1.61E+39

4.56E+47

d)for earth

w1=7.27e-5 rad/s

w2=1.99e-7rad/s

therefore,

L1=I1*w1=9.69*1037kgm2 *7.27e-5 rad/s=70.44e32 kgm2/s

L2=I2*w2=1.336*1047kgm2*1.99e-7rad/s=27.06e39 kgm2/s