Having a hard time solving equation for mass of planet

[lancel916]

This problem is*using*real*data*of*the*first*extrasolar*planet*discovered*around *a normal*star*to*calculate*the mass of the planet. I am really stumped on this equation.I put it as an attachment because I had a hard time entering it by keyboard. So far I figured out from graphs and conversions what k, m & P are equal to.

k = 62.5 m/s
m = 2.11 x 10³⁰ kg
P = 379440 seconds

But when I get to solving the problem I don’t know where to begin. 62.5 m/s X 2.11 x10³⁰ kg I thought but that's Milliseconds times kilograms. also after that do I change the fraction in parentheses to a decimal and times that by the answer to the beginning part? It is an eight part question and this equation is part 4. If someone could explain how this equation is solved I know I can figure all the other similar equations out. It is an online astronomy class which is why I cannot just go to my instructor and ask. Thank you in advance.

 

[!)("/#]

I had a hard time trying to understand what do you want to find, also can you be more specific about the data that you have? Define k and P please.

 

[lancel916]

Here is a shot of question and graph. I figured out what steps to take but I have some questions on some of them.. I don't know how to calculate to a fractional power like 2/3 or 1/3 power of.

Calculate M^(2/3) = (Result 1)

Multiply K times Result 1 (which calculates K * M^(2/3)) = (Piece A)

Divide P by (4.19 x 10^-10) = (Result 2)

Raise Result 2 to the (1/3) power (which is essentially calculating (P/(4.19 x 10^-10))^(1/3)) = (Piece B)

Multiply: (Piece A)*(Piece B) = K * M^(2/3)*(P/(4.19 x 10^-10))^(1/3)) = mp sin(i)

 

[lancel916]

I should have probably given you this information too.

The relationship between the mass of the unseen planet in kilograms, the semiamplitude K in meters per second, and the period P in seconds

This equation can be solved to give mp sin i. (Remember, the mass of the planet is usually expressed in terms of mp sin i because i, the orbital inclination, is often not known. Only if i can be determined will we be able to obtain the planetary mass mp directly.) The mass of the star M can be found in a variety of ways, but that is beyond the scope of this lab. For the exercises that follow, M, P, and K can be read from the Doppler wobble plots.

 

[!)("/#]

Im not sure about this, maybe the problem is too easy or i have no idea about what I am doing, but you have one incognit and one ecuation defined in the following way:

$m_p sin i = K M^{2/3} ( \frac{P}{4.19 \times 10^{-10} \frac{Nm^2}{Kg^2} } )^{1/3}$ (1)

Also you have:$K = 62.5 m/s$

$P = 379440 s$

$M = 2.11 \times 10^{30} Kg$

So puting the data in the ecuation (1):

$m_p sin i = (62.5 m/s) (2.11 \times 10^{30} Kg)^{2/3} ( \frac{379440 s}{4.19 \times 10^{-10} \frac{Nm^2}{Kg^2} } )^{1/3}$

 

[cepheid]

Here is a shot of question and graph. I figured out what steps to take but I have some questions on some of them.. I don't know how to calculate to a fractional power like 2/3 or 1/3 power of.

Basically $x^{\frac{1}{3}} = \sqrt[3]{x}$, and, by the rules of exponents, $x^{\frac{2}{3}} = (x^2)^{\frac{1}{3}} = \sqrt[3]{x^2}$. However, you don't really need to know these definitions to compute the answer. A scientific calculator or a computer will allow you to calculate the result of any number raised to the power of any other number. That's what you would do here.

By the way, how did you get such specific values for k and P? I mean, glancing at the graph, I immediately see that the period is approx. 100 hrs = 360,000 s. But I don't know how you'd get such a precise value. Same with k. I can see that the radial velocity amplitude is about 60 m/s, by looking at the plot. I don't know how you could eyeball it to a precision of 0.5 m/s when the tick mark spacing is 5 m/s.

Calculate M^(2/3) = (Result 1)

Multiply K times Result 1 (which calculates K * M^(2/3)) = (Piece A)

Divide P by (4.19 x 10^-10) = (Result 2)

Raise Result 2 to the (1/3) power (which is essentially calculating (P/(4.19 x 10^-10))^(1/3)) = (Piece B)

Multiply: (Piece A)*(Piece B) = K * M^(2/3)*(P/(4.19 x 10^-10))^(1/3)) = mp sin(i)

Yeah, that sounds about right. Did you have any other specific questions about the computation?

 

[cepheid]

Im not sure about this, maybe the problem is too easy or i have no idea about what I am doing, but you have one incognit and one ecuation defined in the following way:

$m_p sin i = K M^{2/3} ( \frac{P}{4.19 \times 10^{-10} \frac{Nm^2}{Kg^2} } )^{1/3}$ (1)

Also you have:


$K = 62.5 m/s$

$P = 379440 s$

$M = 2.11 \times 10^{30} Kg$

So puting the data in the ecuation (1):

$m_p sin i = (62.5 m/s) (2.11 \times 10^{30} Kg)^{2/3} ( \frac{379440 s}{4.19 \times 10^{-10} \frac{Nm^2}{Kg^2} } )^{1/3}$

The solution i got using Wolfram Mathematica is:

$m_p sin i \approx 9.94744 \times 10^{26} Kg$

That after all seems to be "coherent".

Also for the second item the only thing you have to do is to get the result divided with the mass of jupiter. I got:

$m_p sin i = 0.52355 M_j$

Please read the site rules. We don't provide complete solutions to homework problems on this site.

 

[!)("/#]

Please read the site rules. We don't provide complete solutions to homework problems on this site.

I'm sorry for the inconviniences. I will edit my post.

 

[lancel916]

When you click on the graph it tells you the specifics of where you click. you just have to click within a margin of what they want. which I missed on K. K actually equals 55 m/s. Luckily they let you keep trying till you get it right or give up and submit it. So I just have to rework the problem and it should be right.