Kepler\'s Laws : A new asteroid is discovered which is seen in opposition once e
ID: 158110 • Letter: K
Question
Kepler's Laws :
A new asteroid is discovered which is seen in opposition once every 1.8 years.
(a) What is the sidereal orbital period of the asteroid, in years? For simplicity, you may assume its orbit is circular.
(b) What is the length a of the semimajor axis of the asteroid’s orbit, in AU?
(c) The eccentricity of the orbit of the asteroid is e = 0.1; does the asteroid ever come within 1 AU of the Sun?
Gravity/Orbits
(a) Calculate the ratio gS/gM of the acceleration due to gravity on the Earth’s surface (gS) to the acceleration experienced by the Moon due to the Earth’s gravity (gM).
(b) Calculate the ratio PS/PM of the orbital period of an object orbiting the Earth just above the Earth’s surface (PS) to the Moon’s orbital period (PM). You can make the simplifying assumption that the Earth is much more massive than any satellite, so M + Msat M.
Explanation / Answer
a) siederal orbital period :- The formula to calculate the sideral period is 1/p= 1/E-1/S
p is the sideral period and s is the synodic period
E is 1
1/P= 1-1/1.8=0.8/1.8= 0.45 is the sideral period
b) semimajor axis :- T^2 = R^3 where R = semi-major axis in AU and T = period in years.
1.8^2=R^3
3.24= R^3
R= 1.49 AU is the answer
c) it is hard to reach a AU of 1 with e as 0.1 . because the orbit is of 1.8 years and for that orbit the planet must be ata greater distance from the sun.
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