Question

A planet of mass m 6.75 x 1024 kg is orbiting in a circular path a star of mass M 2.75 x 1029 kg. The radius of the orbit is R 8.05 x107 km. What is the orbital period (in Earth days) of the planet Tplanet?

Answer 1

Answer:

The orbital period of the planet is 387.62 days.

Explanation:

Given that,

Mass of planet$m =6.75\times10^{24}\ kg$

Mass of star $m'=2.75\times10^{29}\ kg$

Radius of the orbit$r =8.05\times10^{7}\ km$

Using centripetal and gravitational force

The centripetal force is given by

$F = \dfrac{mv^2}{r}$

$F=m\omega^2r$

We know that,

$\omega=\dfrac{2\pi}{T}$

$F=m(\dfrac{2\pi}{T})^2r$....(I)

The gravitational force is given by

$F = \dfrac{mm'G}{r^2}$....(II)

From equation (I) and (II)

$m(\dfrac{2\pi}{T})^2r=\dfrac{mm'G}{r^2}$

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

$T^2=\dfrac{4\pi^2R^3}{m'G}$

$T^2=\dfrac{4\times(3.14)^2\times(8.05\times10^{7})^3}{2.75\times10^{29}\times6.67\times10^{-11}}$

$T=2\times3.14\times\sqrt{\dfrac{(8.05\times10^{10})^3}{2.75\times10^{29}\times6.67\times10^{-11}}}$

$T =3.34\times10^{7}\ s$

$T= 387.62\days$

Hence, The orbital period of the planet is 387.62 days.

Answer 2

The planet's orbital period is about 388 days

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Further explanation

Centripetal Acceleration can be formulated as follows:

$\large {\boxed {a = \frac{ v^2 } { R } }$

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:

$\large {\boxed {F = m \frac{ v^2 } { R } }$

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:

$F = ma$

$G \frac{ Mm}{R^2}=m \omega^2 R$

$G M = \omega^2 R^3$

$\frac{GM}{R^3} = \omega^2$

$\omega = \sqrt{ \frac{GM}{R^3}}$

$\frac{2\pi}{T} = \sqrt{ \frac{GM}{R^3}}$

$T = 2\pi \sqrt {\frac{R^3}{GM}}$

$T = 2 \pi \sqrt {\frac{(8.05 \times 10^{10})^3}{6.67 \times 10^{-11} \times 2.75 \times 10^{29}}}$

$T \approx 3.35 \times 10^7 \texttt{ seconds}$

$T \approx 388 \texttt{ days}$

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Learn more

• Impacts of Gravity : brainly.com/question/5330244
• Effect of Earth’s Gravity on Objects : brainly.com/question/8844454
• The Acceleration Due To Gravity : brainly.com/question/4189441

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Answer details

Grade: High School

Subject: Physics

Chapter: Circular Motion

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Keywords: Gravity , Unit , Magnitude , Attraction , Distance , Mass , Newton , Law , Gravitational , Constant