Question

The asteroid Icarus orbits the sun like other planets. Its period is about 410 days. What is its mean distance from the sun

Answer 1

Answer:

Mean distance = 1.61 x 10^8 km

Explanation:

Given the following data;

Orbital period for Icarus, T2 = 410 days

To find the mean distance of Icarus, we would use Kepler's third law of motion.

According to Kepler's third law of planetary motion, the square of any planetary body's orbital period (P) is directly proportional to the cube of its orbit's semi-major axis.

Mathematically, it is given by the formula;

$(\frac {T_{1}}{T_{2}})^2 = (\frac {r_{1}}{r_{2}})^3$

Where;

T1 & T2 is the orbital period of a planetary object.

r1 & r2 is the mean distance of a planetary object.

Also, we know that the orbital period for earth, T1 = 365 days

Mean distance of earth = 1.49x10^8 km

Substituting into the equation, we have;

$(\frac {365}{410})^2 = (\frac {1.49x10^{8}}{r_{2}})^3$

$(\frac {365}{410}})^2 = (\frac {1.49x10^{8}}{r_{2}})^3$

$(0.8902)^2 = (\frac {1.49x10^{8}}{r_{2}})^3$

$(0.7925) = (\frac {1.49x10^{8}}{r_{2}})^3$

Cross-multiplying, we have;

$(r_{2})^3 = \frac {1.49x10^{8}}{0.7925}$

Taking the cube root of both sides;

$r_{2} = 1.61 * 10^8 km$