Answer:
The orbital period of the planet is 387.62 days.
Explanation:
Given that,
Mass of planet[tex]m =6.75\times10^{24}\ kg[/tex]
Mass of star [tex]m'=2.75\times10^{29}\ kg[/tex]
Radius of the orbit[tex]r =8.05\times10^{7}\ km[/tex]
Using centripetal and gravitational force
The centripetal force is given by
[tex]F = \dfrac{mv^2}{r}[/tex]
[tex]F=m\omega^2r[/tex]
We know that,
[tex]\omega=\dfrac{2\pi}{T}[/tex]
[tex]F=m(\dfrac{2\pi}{T})^2r[/tex]....(I)
The gravitational force is given by
[tex]F = \dfrac{mm'G}{r^2}[/tex]....(II)
From equation (I) and (II)
[tex]m(\dfrac{2\pi}{T})^2r=\dfrac{mm'G}{r^2}[/tex]
Where, m = mass of planet
m' = mass of star
G = gravitational constant
r = radius of the orbit
T = time period
Put the value into the formula
[tex]T^2=\dfrac{4\pi^2R^3}{m'G}[/tex]
[tex]T^2=\dfrac{4\times(3.14)^2\times(8.05\times10^{7})^3}{2.75\times10^{29}\times6.67\times10^{-11}}[/tex]
[tex]T=2\times3.14\times\sqrt{\dfrac{(8.05\times10^{10})^3}{2.75\times10^{29}\times6.67\times10^{-11}}}[/tex]
[tex]T =3.34\times10^{7}\ s[/tex]
[tex]T= 387.62\days[/tex]
Hence, The orbital period of the planet is 387.62 days.
The planet's orbital period is about 388 days
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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:
mass of the planet = m = 6.75 × 10²⁴ kg
mass of the star = M = 2.75 × 10²⁹ kg
radius of the orbit = R = 8.05 × 10⁷ km = 8.05 × 10¹⁰ m
Unknown:
Orbital Period of planet = 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]
[tex]T = 2 \pi \sqrt {\frac{(8.05 \times 10^{10})^3}{6.67 \times 10^{-11} \times 2.75 \times 10^{29}}}[/tex]
[tex]T \approx 3.35 \times 10^7 \texttt{ seconds}[/tex]
[tex]T \approx 388 \texttt{ days}[/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
