Question

The planet Mercury travels in an elliptical orbit with eccentricity 0.206. Its minimum distance from the sun is 4.6 ⨯ 107 km. Find its maximum distance from the sun. (If you enter your answer in scientific notation, round the decimal value to two decimal places. Use equivalent rounding if you do not enter your answer in scientific notation.)

Answer 1

Answer:

$a=5.16\times 10^{7}\ km$

Explanation:

It is given that,

The planet Mercury travels in an elliptical orbit with eccentricity 0.206, e = 0.206

The minimum distance from the sun, $b=4.6\times 10^7\ km=4.6\times 10^{10}\ m$

The relation between the minimum and the maximum distance from the sun is given by :

$e=\sqrt{1-\dfrac{b^2}{a^2}}$

a is the maximum distance from the sun

$a^2=\dfrac{b^2}{1-e}$

$a^2=\dfrac{(4.6\times 10^{10})^2}{1-0.206}$

$a=5.16\times 10^{10}\ m$

or

$a=5.16\times 10^{7}\ km$

So, the maximum distance from the sun is $5.16\times 10^{7}\ km$. Hence, this is the required solution.