Physics, 03.01.2021 08:40 andreanaapollon7593

a planet has been detected in a circular orbit around the star Rho1 Cancri with an orbital radius equal to 1.65 x 10^10 m. the orbital period of this planet is approximately 14.5 days which is the approximate mass of the star pho1 cancri

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Answer from: peggycab4201

Approximately $1.69 \times 10^{30}\; \rm kg$ .

Explanation:

Deduction of the formula

Let and denote the mass of the star and the planet, respectively.

Let denote the constant of universal gravitation ( $G \approx 6.67408 \times 10^{-11}\; \rm m^{3} \cdot kg^{-1}\cdot s^{-2}$ .)

Let denote the orbital radius of this planet (assuming that $r\!$ is constant.) The question states that $r = 1.65 \times 10^{10}\; \rm m$ .

The size of gravitational attraction of the star on this planet would be: $\displaystyle \frac{G \cdot M \cdot m}{r^{2}}$ .

If attraction from the star is the only force on this planet, the net force on this planet would be $\displaystyle \frac{G \cdot M \cdot m}{r^{2}}$ .

Let $\omega$ denote the angular velocity of this planet as it travels along its circular orbit around the star. The size of $\omega\!$ could be found from the period of each orbit: $\omega = (2\, \pi) / T$ .

In other words, this planet of mass is in a circular motion with radius and angular velocity $\omega$ . Therefore, the net force on this planet should be equal to $m \cdot \omega^2 \cdot r$ .

Hence, there are two expressions for the net force on this planet:

$\text{Net Force} = \displaystyle \frac{G \cdot M \cdot m}{r^{2}}$ from universal gravitation, and $\displaystyle \text{Net Force} = m \cdot \omega^2 \cdot r = {\left(\frac{2\pi}{T}\right)}^{2} m \cdot r$ from circular motion.

Equate the right-hand side of these two equations:

$\displaystyle \frac{G \cdot M \cdot m}{r^2} = {\left(\frac{2\pi}{T}\right)}^{2}\, m \cdot r$ .

Simplify this equation and solve for , the mass of the star:

$\displaystyle M = \frac{{(2\pi / T)}^2 \cdot r^3}{G}$ .

Notice that , the mass of the planet, was eliminated from the equation. That explains why this question could be solved without knowing the exact mass of the observed planet.

Actual Calculations

Convert the orbital period of this star to standard units:

$\begin{aligned}T &= 14.5\; \text{day} \times \frac{24\; \text{hour}}{1\; \text{day}} \times \frac{3600\; \text{second}}{1\; \text{hour}} \\ & = 1.2528 \times 10^{6}\; \rm \text{second}\end{aligned}$ .

Calculate the mass of the star:

$\begin{aligned} M &= \frac{{(2\pi / T)}^2 \cdot r^3}{G} \\ &\approx \frac{\displaystyle {\left(\frac{2\pi}{1.2528 \times 10^{6}\; \rm s}\right)}^{2} \times \left(1.65 \times 10^{10}\; \rm m\right)^{3}}{6.67408 \times 10^{-11}\; \rm m^{3}\cdot kg^{-1} \cdot s^{-2}}\\ &\approx 1.69 \times 10^{30}\; \rm kg\end{aligned}$ .

Answer from: Quest

from

explanation:

wind is defined by the direction it blows from. for example, a wind blowing from north to south is called a northerly wind.

Answer from: Quest

answer would be letter choice (a), a ball resting on a shelf. would be another example of potential energy > stored energy > not moving.

a ball spinning on a finger > kinetic energy > moving energy

a ball rolling on the ground > kinetic energy > moving energy

a ball flying through the air > kinetic energy > moving energy

hope that : )

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