Question

The planet in a distant solar system has a diameter equal to 1.90×10^7 m

and a free-fall acceleration on the surface that is equal to 12.8 m/s^2
The planet orbits a distance 2.10×10^11 m
from its star (center-to-center distance) with a period of 402 earth days.

What is the mass of the planet?
What is the mass of the star?

Answer 1

(a) 1.73 × 10²⁵ kg.

(b) 4.54 × 10³⁰ kg

Step-by-step explanation:

To address the question regarding the mass of the planet and its star in a distant solar system, we will apply principles of gravitational physics and orbital mechanics.

We are given:

• d = 1.90 × 10⁷ m
• g = 12.8 m/s²
• R = 2.10 × 10¹¹ m
• T = 402 Earth days

We want to find:

(a) mass of the planet

(b) mass of the star

`\hrulefill`

(a) Determining the Mass of the Planet
`\hrulefill`

The free-fall acceleration on the surface of a planet is given by the formula:

`g = (Gm_p)/(R^2)`

Where:

• 'g' is the acceleration due to gravity
• 'G' is the gravitational constant (6.67 × 10⁻¹¹ Nm²/kg²)
• 'm_p' is the mass of the planet (which we need to find)
• 'R' is the radius of the planet (in meters)

We can rearrange the formula to solve for 'm_p':

`\Longrightarrow m_p = (R^2g)/(G)`

Plug in what is given:

`\Longrightarrow m_p = \frac{\left(\frac{1.90 * 10^7 \text{ m}}{2}\right)^2(12.8 \text{ m/s$^2$})}{6.67 * 10^(-11) \frac{\text{Nm}^2}{\text{kg}^2}}\\\\\\\\\therefore m_p \approx \boxed{1.73 * 10^(25) \text{ kg}}`

Thus, the mass of the planet is approximately 1.73 × 10²⁵ kg.

`\hrulefill`

(b) Determining the Mass of the Star
`\hrulefill`

To find the mass of the star, we use Kepler's Third Law (assuming circular orbit), which relates the orbital period of a planet to its distance from the star. The formula is:

`T^2=(4 \pi^2 R^3)/(G m_p)`

Where:

• 'T' is the period (in seconds)
• 'R' is the orbit radius (in meters)
• 'G' is the gravitational constant
• 'm_p' is the mass of the planet (in this case the star)

We can rearrange the formula to solve for 'm_p':

`\Longrightarrow m_p=(4 \pi^2 R^3)/(G T^2)`

Plug in what is given:

`\Longrightarrow m_p=\frac{4 \pi^2 (2.10 * 10^(11) \text{ m})^3}{(6.67 * 10^(-11) \frac{\text{Nm}^2}{\text{kg}^2})\left(\frac{402 \text{ days}}{}*\frac{24 \text{ hr}}{1 \text{ day}}* \frac{3600 \text{ s}}{1 \text{ hr}}\right)^2}\\\\\\\\\therefore m_p \approx \boxed{4.54 * 10^(30) \text{ kg}}`

Thus, the mass of the star is approximately 4.54 × 10³⁰ kg.

Answer 2

`\sf m_{\textsf{planet}} \approx 1.73 * 10^(25) \, \textsf{kg}`

`\sf m_{\textsf{star}} \approx 4.54 * 10^(30) \, \textsf{kg}`

Step-by-step explanation:

Planet and Star Mass Calculations:

Planet Mass:

1. Radius calculation:

Diameter =
`\sf 1.90 * 10^7 \, \textsf{m}`

Radius = Diameter / 2 =
`\sf 9.5 * 10^6 \, \textsf{m}`

2. Gravitational acceleration:

`\sf g = 12.8 \, \textsf{m/s}^2`

3. Gravitational constant:

`\sf G = 6.6743 * 10^(-11) \, \textsf{Nm}^2/\textsf{kg}^2`

4. Planet mass formula:

`\sf m_{\textsf{planet}} = (g \cdot R^2)/(G)`

Substitute the known value:

`\sf m_{\textsf{planet}} = (12.8 \cdot (9.5 * 10^6)^2)/(6.6743 * 10^(-11))`

`\sf m_{\textsf{planet}} \approx 1.73 * 10^(25) \, \textsf{kg}`

Therefore, the planet's mass is approximately
`\sf 1.73 * 10^(25) \, \textsf{kg}.`

Star Mass:

1. Orbital distance:

Distance =
`\sf 2.10 * 10^(11) \, \textsf{m}`

2. Orbital period:

Period =
`\sf 402` Earth days

Convert to seconds: Period =
`\sf 402` days
`\sf *`
`\sf 24` hours/day
`\sf *`
`\sf 3600` seconds/hour =
`\sf 35,136,000` seconds

3. Orbital velocity:

`\sf v = \frac{2\pi \cdot \textsf{distance}}{\textsf{period}}`

`\sf v = \frac{2\pi \cdot 2.10 * 10^(11) \, \textsf{m}}{35,136,000 \, \textsf{seconds}}`

`\sf v \approx 37.7 \, \textsf{m/s}`

4. Star mass formula (assuming circular orbit):

`\sf m_{\textsf{star}} = \frac{G \cdot m_{\textsf{planet}} \cdot \textsf{period}^2}{4\pi^2 \cdot \textsf{distance}^3}`

Substitute the known value:

`\sf m_{\textsf{star}} = (6.6743 * 10^(-11) \cdot 1.73 * 10^(25) \cdot (35,136,000)^2)/(4\pi^2 \cdot (2.10 * 10^(11))^3)`

`\sf m_{\textsf{star}} \approx 4.54 * 10^(30) \, \textsf{kg}`

Therefore, the star's mass is approximately
`\sf 4.54 * 10^(30) \, \textsf{kg}.`

Note: These are approximate values due to rounding and simplifying calculations.