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Consider as a system the Sun with Saturn in a circular orbit around it. Find the magnitude of the change in the velocity of the Sun relative to the center of mass of the system during the time Saturn completes half an orbit. Assume the mass of the Sun is 5.68 x10^29 kg, the mass of Saturn is 5.68 x10^26 kg, its period is 9.29 x10^8 s, and the radius of its orbit is 1.43 x 10^12 m. Ignore the influence of other celestial objects.?

Respuesta :

davidlcastiblanco

Answer:

[tex]v_{su} = 19.44 m/s[/tex]

Explanation:

[tex]m_{su}=5.68x10^{29}kg\\m_{sa}=5.68x10^{26}kg[/tex]

[tex]T=9.29x10^8\\r_{o}=1.43x10^{12}[/tex]

If the sun considered as x=0 on the axis to put the center of the mass as a:

[tex]m_{su}*r_{o}=(m_{sa}+m_{su})*r_{1}[/tex]

solve to r1

[tex]r_1=\frac{m_{sa}*r_{o}}{m_{sa}+m_{su}}=\frac{5.68x10^{26}*1.43x10^{12}}{5.68x10^{26}+5.68x10^{26}}[/tex]

[tex]r_1=1.428x10^9m[/tex]

Now convert to coordinates centered on the center of mass.  call the new coordinates x' and y' (we won't need y').  Now since in the sun centered coordinates the angular momentum was  

[tex]L = \frac{m_{sa}*2*pi*r_1^2}{T}[/tex]

where T = orbital period

then L'(x',y') = L(x) by conservation of angular momentum.  So that means

[tex]L_{sun}=\frac{m_{sa}*2*\pi *( 2r_{o}*r_1 -r_1^2)}{T}[/tex]

Since

[tex]L_{su}= m_{su}*v_{su}*r_1[/tex]

then

[tex]v_{su}=\frac{m_{sa}*2*pi*(2r_{o}*r_{1}-r_{1}^2)}{T*m_{sa}*r_1}[/tex]

[tex]v_{su} = 19.44 m/s[/tex]

onyebuchinnaji

The magnitude of the change in the velocity of the Sun relative to the center of mass of the system is [tex]4.83 \times 10^{3} \ m/s[/tex].

The given parameters;

  • mass of the Sun, m₁ = 5.68 x 10²⁹ kg
  • mass of the Saturn, m₂ = 5.68 x 10²⁶ kg
  • period of Saturn, T = 9.29 x 10⁸ s
  • radius of Saturn, r = 1.43 x 10¹² m.

The center mass of two Planets is calculated as follows, assuming Sun at the reference 0 point;

[tex]r_{c} = \frac{m_1r_1 \ + m_2 r_2}{m_1 + m_2} \\\\r_{c} = \frac{0 \ + \ 5.68\times 10^{26} \times 1.43 \times 10^{12} }{5.68 \times 10^{26} + 5.68 \times 10^{29}} \\\\r_{c} = 1.43 \times 10^9 \ m[/tex]

The angular velocity of the Saturn when it completes half orbit is calculated as;

[tex]\omega = \frac{\pi}{T} = \frac{3.142}{9.29 \times 10^8} \\\\\omega = 3.38 \times 10^{-9} \ rad/s[/tex]

The angular momentum of the center mass;

[tex]L_{sa} = I \omega\\\\L_{sa} = Mr_c^2 \omega\\\\L_{sa} = (5.68\times 10^{26} )(1.43\times 10^{12})^2\times 3.38 \times 10^{-9}\\\\L_{sa} = 3.925 \times 10^{42} \ kgm^2[/tex]

Apply the principle of conservation of angular momentum to determine the velocity of the sun relative to the center mass;

[tex]I_s \omega_s = 3.925\times 10^{42}\\\\m_s r_c^2 (\frac{v_s}{r_c} )= 3.925\times 10^{42}\\\\v_s = \frac{ 3.925\times 10^{42}}{m_s r_c^2} \\\\v_s = \frac{ 3.925\times 10^{42}}{(5.68\times 10^{29})\times (1.43 \times 10^9)}\\\\v_s = 4.83 \times 10^3 \ m/s[/tex]

Thus, the magnitude of the change in the velocity of the Sun relative to the center of mass of the system is [tex]4.83 \times 10^{3} \ m/s[/tex]

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