Question

a) An asteroid revolves around the Sun with a mean orbital radius 2.5 times that of Earth’s. Predict the period of the asteroid in Earth years.

(Show all work, including formulas and units for full credit.)

b) Saturn requires 29 years to circle the Sun. Find Saturn’s average distance from the Sun as a multiple of Earth’s average distance from the Sun.

(Show all work, including formulas and units for full credit.)

c) The Moon has a period of 27.3 days and a mean distance of 3.9×10⁵ km from its center to the center of Earth.

i) Use Kepler’s laws to find the period of a (hypothetical) satellite in orbit 8.20×10³ km from the center of Earth.

ii) How far above Earth’s surface is this satellite (Hint: Look up the Earth's diameter online, convert it to the radius, and also to kilometers. Keep in mind that the satellite rotates from the center of the Earth, and the question asks for the height above the surface of the Earth)?

Answer 1

a)

T_e = Time period of earth for orbit around the sun = 1 yr

R_e = Radius of earth for orbit around the sun = R

T_a = Time period of asteroid for orbit around the sun = ?

R_a = Radius of asteroid for orbit around the sun = (2.5) R

We know from Kepler's third law that

T² R³

hence , for comparing the time period and orbital radius of earth and asteroid , we have

T_a² /T_e² = R_a³ /R_e³

T_a² /(1)² = ((2.5) R)³ /R³

T_a² = (2.5)³ R³ /R³

T_a² = (2.5)³

T_a = 3.95

So time period is 3.95 earth years

b)

T_e = Time period of earth for orbit around the sun = 1 yr

R_e = Radius of earth for orbit around the sun = R

T_s = Time period of saturn for orbit around the sun = 29 yrs

R_s = Radius of saturn for orbit around the sun = ?

We know from Kepler's third law that

T² R³

hence , for comparing the time period and orbital radius of earth and asteroid , we have

T_s² /T_e² = R_s³ /R_e³

(29)² /(1)² = R_s³ /R³

R_s³ = (29)² R³

R_s = (29)^2/3 R

R_s = 9.44 R

So 9.44 times the earth's distance

c)

i)

T_m = Time period of moon for orbit around the earth = 27.3 days

T_s = Time period of satellite for orbit around the earth = ?

R_m = center to center distance between moon and earth = 3.90 x 10⁵ km = 3.90 x 10⁸ m

R_s = center to center distance between satellite and earth = 8.20 x 10³ km = 8.20 x 10⁶ m

Using the equation

T_s² /T_m² = R_s³ /R_m³

T_s² /(27.3)² = (8.20 x 10⁶)³ /(3.90 x 10⁸)³

T_s = 0.0832 days

ii)

R = radius of earth = 6371 km

R_s = center to center distance between satellite and earth = 8.20 x 10³ km = 8200 km

h = height of the satellite above the surface of earth

h = R_s - R

h = (8200) - (6371)

h = 1829 km