Answer: [tex]1.079(10)^{-6}Earth-years[/tex]
Explanation:
This problem can be solved using the Third Kepler’s Law of Planetary motion:
“The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”.
This law states a relation between the orbital period [tex]T[/tex] of a body (the asteroid in this case) orbiting a greater body in space (the Sun in this case) with the size [tex]a[/tex] of its orbit:
[tex]T^{2}=\frac{4\pi^{2}}{GM}a^{3}[/tex] (1)
Where;
[tex]G[/tex] is the Gravitational Constant and its value is [tex]6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}[/tex]
[tex]M=1.989(10)^{30}kg[/tex] is the mass of the Sun
[tex]a=2.47(Earth-radius)=2.47(6371000m)=15736370m[/tex] is orbital radius of the orbit the asteroid describes around the Sun.
Now, if we want to find the period, we have to express equation (1) as written below and substitute all the values:
[tex]T=\sqrt{\frac{4\pi^{2}}{GM}a^{3}}[/tex] (2)
[tex]T=\sqrt{\frac{4\pi^{2}}{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(1.989(10)^{30}kg)}(15736370m)^{3}}[/tex] (3)
[tex]T=34.0428s[/tex] (4) This is the period in seconds, but we were asked to find it in Earth years. So, we have to make the conversion:
[tex]T=34.0428s.\frac{1Earth-hour}{3600s}.\frac{1Earth-day}{24Earth-hour}.\frac{1Earth-year}{365Earth-day}[/tex]
Finally:
[tex]T=1.079(10)^{-6}Earth-years[/tex] This is the period of the asteroid around the Sun in Earth years.
The asteroid's orbital period is about 3.88 years
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Centripetal Acceleration can be formulated as follows:
[tex]\large {\boxed {a = \frac{ v^2 } { R } }[/tex]
a = Centripetal Acceleration ( m/s² )
v = Tangential Speed of Particle ( m/s )
R = Radius of Circular Motion ( m )
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Centripetal Force can be formulated as follows:
[tex]\large {\boxed {F = m \frac{ v^2 } { R } }[/tex]
F = Centripetal Force ( m/s² )
m = mass of Particle ( kg )
v = Tangential Speed of Particle ( m/s )
R = Radius of Circular Motion ( m )
Let us now tackle the problem !
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Given:
Orbital Radius of Earth = R
Orbital Radius of Asteroid = R' = 2.47 R
Orbital Period of Earth = T = 1 year
Unknown:
Orbital Period of Asteroid = T' = ?
Solution:
Firstly , we will use this following formula to find the orbital period:
[tex]F = ma[/tex]
[tex]G \frac{ Mm}{R^2}=m \omega^2 R[/tex]
[tex]G M = \omega^2 R^3[/tex]
[tex]\frac{GM}{R^3} = \omega^2[/tex]
[tex]\omega = \sqrt{ \frac{GM}{R^3}}[/tex]
[tex]\frac{2\pi}{T} = \sqrt{ \frac{GM}{R^3}}[/tex]
[tex]T = 2\pi \sqrt {\frac{R^3}{GM}}[/tex]
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Next , we could find the asteroid's orbital period by using above formula:
[tex]T' : T = 2\pi\sqrt{ \frac{(R')^3}{GM}} : 2\pi\sqrt{ \frac{R^3}{GM}}[/tex]
[tex]T' : T = \sqrt{(R')^3} : \sqrt{R^3}[/tex]
[tex]T' : 1 = \sqrt{(2.47R)^3} : \sqrt{R^3}[/tex]
[tex]T' : 1 = \sqrt{2.47^3}\sqrt{R^3} : \sqrt{R^3}[/tex]
[tex]T' : 1 = \sqrt{2.47^3} : 1[/tex]
[tex]T' = \sqrt{2.47^3}[/tex]
[tex]T' \approx 3.88 \texttt{ years}[/tex]
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Grade: High School
Subject: Physics
Chapter: Circular Motion
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Keywords: Gravity , Unit , Magnitude , Attraction , Distance , Mass , Newton , Law , Gravitational , Constant
