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Chapter 14: Problem 36

The orbit of the Earth about the Sun is almost circular. The closest and farthest distances are \(1.47 \times 10^{8} \mathrm{~km}\) and \(1.52 \times\) \(10^{8} \mathrm{~km}\), respectively. Determine the maximum variations in ( \(a\) ) potential energy, (b) kinetic energy, ( \(c\) ) total energy, and (d) orbital speed that result from the changing Earth-Sun distance in the course of 1 year. (Hint: Use conservation of energy and angular momentum.)

Short Answer

Expert verified
This question requires computing different forms of energies and comparing them at different Earth-Sun distances. The process demands the computational knowledge in physics, understanding of the conservation of energy, gravitational potential energy and conservation of angular momentum combined with kinetic energy equations.

Step by step solution

01

Calculate potential energy

To find the potential energy, use the formula for gravitational potential energy: \[\text{Potential Energy (PE)} = - \frac{GMm}{r}\]whereG = gravitational constant = \(6.67 \times 10^{-11} \, \text{m}^3 \, \text{kg}^{-1} \, \text{s}^{-2}\),M = solar mass = \(1.99 \times 10^{30} \, \text{kg}\),m = Earth's mass = \(5.98 \times 10^{24} \, \text{kg}\),r = distance between Earth and the Sun.Evaluate this expression for both minimum and maximum distances to calculate the difference or variation. Be sure to convert kilometers to meters, as G is given in m^3/kg/s^2.
02

Determine kinetic energy

To determine kinetic energy, first remember that total energy (E) is conserved and is given as \[E = \text{KE} + \text{PE} = - \frac{GMm}{2a}\]from which \[\text{KE} = E - \text{PE} = - \frac{GMm}{2a} - (- \frac{GMm}{r}) = \frac{GMm}{2a} - \frac{GMm}{r}\]wherea = semi-major axis = \(\frac{r_{\text{min}} + r_{\text{max}}}{2}.\)Again calculate for both situations (closest and farthest Earth-Sun distances) and find the difference.
03

Calculate total energy

The total energy variation is just the sum of potential and kinetic energy variations.
04

Calculate orbital speed

For the orbital speed, use that the angular momentum is conserved, i.e., \[m v_{\text{min}} r_{\text{min}} = m v_{\text{max}} r_{\text{max}}\]to find \[v_{\text{max can be found from}}\frac{v_{\text{min}}r_{\text{min}}}{r_{\text{max}}}\]and then calculate \(v_{\text{min}}\) using the kinetic energy equation \(KE = \frac{1}{2} m v_{\text{min}}^2.\) Then calculate the difference in speeds.

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