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According to Kepler’s Third Law, the sidereal period, in years, of a planet squared is equal to its semi-major axis, in Astronomical Units, cubed. Which statement is true in the context of this law?

a. A planet that has a greater mass takes more time to orbit the Sun.
b. A planet that has a larger semi-major axis takes more time to orbit the Sun.
c. A planet that has a greater size takes more time to orbit the Sun.
d. A planet that has a larger semi-major axis takes less time to orbit the Sun.
e. A planet that has a longer solar day is farther from the Sun.


Is the answer to this question B?????

Thank you

2006-10-02 07:11:08 · 5 answers · asked by Roman K 2 in Science & Mathematics Astronomy & Space

5 answers

In equation form, Kepler's third law states that:

T^2 = D^3


where T is the orbital period in years
D is the semi-major axis in A.U.


The law says nothing about the mass, size, or length of day of a planet, so answers a, c, and e are obviously incorrect.

Look at the above equation. As D gets larger, then so does D^3. If D^3 gets larger than so does T^2, and therefore so does T.

Answer B is correct.

2006-10-02 07:24:06 · answer #1 · answered by hfshaw 7 · 0 0

Yes, I do think the option 'b' is the right one.

NOTE
____________________________________________________
[Kepler's third law states :
The squares of the orbital periods of planets are directly proportional to the cubes of the semi-major axis of the orbits.
T^2 proportional to a^3
where,
T = orbital period of planet
a = semimajor axis of orbit ]

2006-10-02 07:20:01 · answer #2 · answered by Innocence Redefined 5 · 0 0

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2016-10-15 10:49:15 · answer #3 · answered by ? 4 · 0 0

Yup, B is the only right answer.

2006-10-02 08:16:53 · answer #4 · answered by campbelp2002 7 · 0 0

Hi. B is correct.

2006-10-02 07:21:34 · answer #5 · answered by Cirric 7 · 0 0

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