Chapter 13, Problem 048 The mean distance of an asteroid from the Sun is 1.20 times that of Earth from the Sun. From Kepler\'s law of periods, calculate the number of years required for the asteroid to make one revolution around the Sun. Number Units the tolerance ts +/-296
Solution
As we know that;
From keplers law,
= Time period is proportional to r^(3/2)
= r = distacne of the planet from sun
So,
= Ta/TE = (ra/rE)^(3/2)
= Ta = time period of asteriod
= TE = time period of earth = 1 year
= ra = distacne of asteriod from sun = 1.2rE
= rE = distacne of earth from sun
Then,
= Ta/TE = 1.2^(3/2) = 1.31
= Ta = 1.31*TE
= Ta = 1.31 years
.
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Chapter 13- Problem 048 The mean distance of an asteroid from the Sun.docx
1. Chapter 13, Problem 048 The mean distance of an asteroid from the Sun is 1.20 times that of
Earth from the Sun. From Kepler's law of periods, calculate the number of years required for the
asteroid to make one revolution around the Sun. Number Units the tolerance ts +/-296
Solution
As we know that;
From keplers law,
= Time period is proportional to r^(3/2)
= r = distacne of the planet from sun
So,
= Ta/TE = (ra/rE)^(3/2)
= Ta = time period of asteriod
= TE = time period of earth = 1 year
= ra = distacne of asteriod from sun = 1.2rE
= rE = distacne of earth from sun
Then,
= Ta/TE = 1.2^(3/2) = 1.31
= Ta = 1.31*TE
= Ta = 1.31 years
