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A comet of mass 1.20  1010 kg moves in an elliptical orbit around the Sun. Its distance from the Sun ranges between 0.500 AU and 50.0 AU.
(a) What is the eccentricity of its orbit?
(b) What is its period?
(c) At aphelion what is the potential energy of the comet-Sun system? Note: 1 AU = one astronomical unit = the average distance from Sun to Earth = 1.496  1011 m.

Answer :

Answer:

Part a)

[tex]T = 354.7 years[/tex]

Part c)

[tex]U = -2.12 \times 10^{17} J[/tex]

Explanation:

Part a)

Eccentricity of ellipse is given as

[tex]e = \sqrt{1 - \frac{b^2}{a^2}}[/tex]

here we know that

b = 0.5 AU

a = 50 AU

so we will have

[tex]e = \sqrt{1 - \frac{0.5^2}{50^2}}[/tex]

[tex]e = 0.99[/tex]

Part b)

Time period around SUN is given as

[tex]T = 2\pi\sqrt{\frac{a^3}{GM}}[/tex]

[tex]T = 2\pi\sqrt{\frac{(50\times 1.496 \times 10^{11})^3}{(6.67 \times 10^{-11})(1.98 \times 10^{30})}}[/tex]

[tex]T = 354.7 years[/tex]

Part c)

As we know that potential energy of the system is given as

[tex]U = - \frac{GMm}{r}[/tex]

[tex]U = -\frac{(6.67 \times 10^{-11})(1.20 \times 10^{10})(1.98 \times 10^{30})}{50 \times 1.496 \times 10^{11})}[/tex]

[tex]U = -2.12 \times 10^{17} J[/tex]

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