The average radius r in meters of a planet's path around the sun is given by the formula r=[(3.35 x 10^18)t^2]^1/3, where t is the time in seconds that it takes the planet to go around the sun.
a. Rewrite the equation using the root symbol.
b. Neptune takes about 5.2 x 10^9 seconds to go around the sun. What is the average radius of Neptune's ath around the sun?
c. The time it takes a planet to go around the sun is called a year. Compare Neptune's year with Earth's year. Which planet has the longer year? How much longer is it.?
b. Neptune takes about 5.2 x 10^9 seconds to go around the sun. What is the average radius of Neptune's ath around the sun?
c. The time it takes a planet to go around the sun is called a year. Compare Neptune's year with Earth's year. Which planet has the longer year? How much longer is it.?
b-Since the planet;s period is defined by P = 2Pisqrt(r^3/µ), where r == the planets orbital radius and µ = the gravitational constant of the sun, solving for r yields r = (µP^2/4Pi^2)^(1/3).
Thus,
r = [4.68772x10^21(5.2x10^9)/4Pi^2]^1/3 = 2,794,059,276 miles.
c-Substitute the Earth's mean radius, 92,960,242 miles, and Neptunes mean radius, just calculated, into the expression for period given above, to get the comparison of orbital periods.
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a. To rewrite the equation using the root symbol, we need to express the exponent as a root. The expression [(3.35 x 10^18)t^2]^1/3 can be rewritten as the cube root of [(3.35 x 10^18)t^2].
So, the new equation using the root symbol is r = ∛[(3.35 x 10^18)t^2].
b. Given that Neptune takes about 5.2 x 10^9 seconds to go around the sun, we can substitute this value into the equation to find the average radius. Let's plug in the value and calculate:
r = ∛[(3.35 x 10^18)(5.2 x 10^9)^2]
First, calculate 5.2 x 10^9 squared:
(5.2 x 10^9)^2 = (5.2)^2 x (10^9)^2 = 27.04 x 10^18
Now, let's substitute this value back into the equation:
r = ∛[(3.35 x 10^18)(27.04 x 10^18)]
= ∛(90.544 x 10^36)
To simplify, we can express 90.544 as 9.0544 x 10^1:
r = ∛(9.0544 x 10^1 x 10^36)
= ∛(9.0544 x 10^37)
Now we can calculate the cube root:
r ≈ 4.256 x 10^12 meters
Therefore, the average radius of Neptune's path around the sun is approximately 4.256 x 10^12 meters.
c. To compare Neptune's year with Earth's year, we need to know the time it takes for Earth to go around the sun. Earth's year is approximately 365.25 days, which is equivalent to approximately 31,557,600 seconds.
Now, we can compare the times:
Neptune's year: 5.2 x 10^9 seconds
Earth's year: 31,557,600 seconds
To determine which planet has the longer year, we compare the values. In this case, Earth's year (31,557,600 seconds) is longer than Neptune's year (5.2 x 10^9 seconds).
To calculate how much longer Earth's year is compared to Neptune's, we can find the ratio:
(31,557,600 seconds) / (5.2 x 10^9 seconds)
≈ 6.065
Therefore, Earth's year is approximately 6 times longer than Neptune's year.