How do you calculate the mass of the sun, $M_{sun}$ $M_"sun"$ , using Kepler's third law ( $T^{2} = \frac{4 π^{2} r^{3}}{G M_{sun}}$ $T^2=(4 pi^2 r^3)/(G M_"sun")$ )?

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How do you calculate the mass of the sun, $M_{sun}$ $M_"sun"$ , using Kepler's third law ( $T^{2} = \frac{4 π^{2} r^{3}}{G M_{sun}}$ $T^2=(4 pi^2 r^3)/(G M_"sun")$ )?

Assume the period of the Earth is $T = 3.156 \times 10^{7}$ $T=3.156xx10^7$ seconds and the Earth's distance from the Sun is $1.496 \times 10^{11}$ $1.496xx10^11$ meters.

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Answer 1 · 2017-05-13T02:52:03

Plug the given values into the given equation. Answer: $\approx 1.98955 \times 10^{30}$ $~~1.98955xx10^30$ $kg$ $"kg"$

Explanation:

Since we are given the equation:
$T^{2} = \frac{4 π^{2} r^{3}}{G M_{s u n}}$ $T^2=(4pi^2r^3)/(GM_(sun))$
where $T = 3.156 \times 10^{7}$ $T=3.156xx10^7$ seconds, $r = 1.496 \times 10^{11}$ $r=1.496xx10^11$ meters, $π \approx 3.14$ $pi~~3.14$ is the mathematical constant, and $G = 6.67 \times 10^{- 11}$ $G=6.67xx10^-11$ as the gravitational constant, we can first solve for $M_{s u n}$ $M_(sun)$ with variables then substitute the given values to find $M_{s u n}$ $M_(sun)$ :

First, we will use variables to solve for $M_{s u n}$ $M_(sun)$ to avoid the amount of numbers in the equation:
$T^{2} = \frac{4 π^{2} r^{3}}{G M_{s u n}}$ $T^2=(4pi^2r^3)/(GM_(sun))$
$T^{2} (G M_{s u n}) = 4 π^{2} r^{3}$ $T^2(GM_(sun))=4pi^2r^3$
$M_{s u n} = \frac{4 π^{2} r^{3}}{T^{2} G}$ $M_(sun)=(4pi^2r^3)/(T^2G)$

Now, we can substitute our given values:
$M_{s u n} = \frac{4 π^{2} {(1.496 \times 10^{11})}^{3}}{{(3.156 \times 10^{7})}^{2} (6.67 \times 10^{- 11})}$ $M_(sun)=(4pi^2(1.496xx10^11)^3)/((3.156xx10^7)^2(6.67xx10^-11))$

$\approx 1.98955 \times 10^{30}$ $~~1.98955xx10^30$ $kg$ $"kg"$ rounded to 5 decimals

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How do you calculate the mass of the sun, $M_{sun}$ $M_"sun"$ , using Kepler's third law ( $T^{2} = \frac{4 π^{2} r^{3}}{G M_{sun}}$ $T^2=(4 pi^2 r^3)/(G M_"sun")$ )?

Assume the period of the Earth is $T = 3.156 \times 10^{7}$ $T=3.156xx10^7$ seconds and the Earth's distance from the Sun is $1.496 \times 10^{11}$ $1.496xx10^11$ meters.

1 Answer

Explanation:

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How do you calculate the mass of the sun, MsunM_"sun", using Kepler's third law (T2=4π2r3GMsunT^2=(4 pi^2 r^3)/(G M_"sun"))?

Assume the period of the Earth is T=3.156×107T=3.156xx10^7 seconds and the Earth's distance from the Sun is 1.496×10111.496xx10^11 meters.

1 Answer

Explanation:

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Impact of this question

How do you calculate the mass of the sun, $M_{sun}$ $M_"sun"$ , using Kepler's third law ( $T^{2} = \frac{4 π^{2} r^{3}}{G M_{sun}}$ $T^2=(4 pi^2 r^3)/(G M_"sun")$ )?

Assume the period of the Earth is $T = 3.156 \times 10^{7}$ $T=3.156xx10^7$ seconds and the Earth's distance from the Sun is $1.496 \times 10^{11}$ $1.496xx10^11$ meters.