(a) two hypothetical planets are moving about the sun in elliptical orbits having equal major axes. the minor axis of one, however, is half that of the other. how do the periods of the two planets compare? (b) the moon revolves about the earth in 27.3 days in an elliptical orbit whose semimajor axis is 60 times the earth’s radius. what would be the period of a satellite revolving close to the earth’s surface? (since the satellite is close to the earth’s surface, let its semimajor axis be the earth’s radius, r.)

a) They are equal.

b) 1 hour and 24 minutes.

Step-by-step explanation:

We are going to solve this exercise applying the third law of Kepler.

Third law of Kepler : ''The square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit''

Let be ''T'' the orbital period and ''a'' the semi-major axis ⇒

Where K is a constant.

a)

Both planets have the same period because they have equal semi-major axis ⇒ T1 = T2

Being T1 the period of the first planet and T2 the period of the second planet.

b)

Let be ''r'' the Earth's radius

For the Moon :

For the satellite :

⇒

⇒

T ≅ 1 hour and 24 minutes

answer: 94

step-by-step explanation: 79+91+92+94+94 = 450 divided by 5 (since there are 5 number that are being added) = 90