You are the science officer on a visit to a distant solar system. Prior to landing on a planet you measure its diameter to be 2.070E+7 m and its rotation to be 19.8 hours. You have previously determined that the planet orbits 1.590E+11 m from its star with a period of 368.0 Earth days. Once on the surface you find that the free-fall acceleration is 13.00 m/s2.
What is the mass of the star (in kg)?
I have posted this multiple times and can not figure it out.
Yes, thank you but as I said, I still can not figure out from what you gave me so maybe someone else will explain better (figuring out the mass of the star that is).
Why did the scientist go to the comedy club? To understand the mass of the star, they needed a good laugh!
But let's get serious for a moment. To calculate the mass of the star, we can use Newton's law of universal gravitation, which states that the gravitational force between two objects is given by:
F = (G * m1 * m2) / r^2
Where F is the force of gravity, G is the gravitational constant, m1 and m2 are the masses of the objects, and r is the distance between them.
In this case, we know the distance between the planet and the star (r), the free-fall acceleration (g), and the period of the planet's orbit (T). We can use these values to calculate the mass of the star:
1. Calculate the mass of the planet (mp) using the formula:
mp = g * (2.070E+7 m)^2 / 4π^2
2. Convert the period of the planet's orbit (T) into seconds:
T_seconds = 368.0 days * 24 hours * 60 minutes * 60 seconds
3. Calculate the mass of the star (ms) using the formula:
ms = (4π^2 * (1.590E+11 m)^3) / (G * T_seconds^2)
Remember, G is the gravitational constant, which is approximately 6.67430E-11 m^3 / (kg * s^2).
Give it a try, and don't forget to keep a cosmic sense of humor along the way!
To determine the mass of the star, we can use Newton's law of universal gravitation, which states that the gravitational force between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.
Step 1: Convert the given values to SI units.
- Planet's diameter: 2.070E+7 m
- Planet's rotation: 19.8 hours = 19.8 * 60 * 60 seconds (convert hours to seconds)
- Planet's orbit radius: 1.590E+11 m
- Planet's orbit period: 368.0 Earth days = 368.0 * 24 * 60 * 60 seconds (convert days to seconds)
- Free-fall acceleration: 13.00 m/s²
Step 2: Calculate the planet's mass.
Using the formula for centripetal force, we can solve for the planet's mass (m):
Centripetal force = gravitational force between the star and the planet
(m * v²) / r = (G * M * m) / r²
Here,
m = mass of the planet
v = linear velocity of the planet in its orbit (2πr / T)
r = orbit radius of the planet
G = gravitational constant (6.67430 × 10^-11 m³ kg^-1 s^-2)
M = mass of the star
Step 3: Calculate the linear velocity of the planet.
v = (2πr) / T
Here,
T = orbit period of the planet
Step 4: Calculate the mass of the star.
Using the already derived equation, rearrange it to solve for M:
M = (m * v² * r²) / (G * m)
= (v² * r²) / G
Step 5: Calculate the mass of the star using the given values.
Plug in the values into the equation derived in Step 4 and calculate the mass (M).
Note: Since the units of the given values are already in SI units, we can directly substitute them into the equation.
M = (v² * r²) / G
Remember to convert the mass of the star to its appropriate scientific notation if necessary.
I hope this explanation helps you solve the problem!
The reference I gave you for Kepler's Third Law did not solve the problem for stars other than the sun. A better reference would be
http://www.astro.cornell.edu/academics/courses/astro101/java/Finding%20Exosolar%20Planets.htm
There, you will find the formula.
P^2 = 4 pi^2*r^3/[G*(m1 + m2)]
The author also derives it there.
In your case, ignore the planet's mass m2 because it will be negligible compared to the star's mass, m1.
P is the period, 368 days = 3.18*10^7 s
r is the orbital radius, 1.59*10^11 m.
Look up the value of G and solve for the star's mass m1, in kilograms.
It should be not much different from the sun's mass, because the r and P values you are using are close to those of the Earth orbiting the sun.
I have answered it once and will not do it again.
http://www.jiskha.com/display.cgi?id=1322287372