This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. Communication satellites are placed in a geosynchronous orbit, i. e., in a circular orbit such that they complete one full revolution about the earth in one sidereal day (23.934 h), and thus appear stationary with respect to the ground. Determine the altitude of these satellites above the surface of the earth in both SI and U. S. customary units. The altitude is SI units in km. The altitude is U. S. customary units in mi.

the altitude of these satellites above the surface of the earth;

35790 km ( SI units )

22243.63 miles ( U.S. customary unit )

Explanation:

Given the data in the question;

Time taken by the satellite to complete on revolution is 23.934 hours

= 23.934 × 60 × 60 = 86162.4 seconds

now, let h represent altitude, r represent Orbit radius, v represent Orbit speed.

we know that

v² = GM/r

= gR²/r

= (9.81m/s² × ( 6.37 × 10⁶ m)²) / r

v = 19.95 × 10⁶ / √r   let this be equation

also;

time t = 2πr/v

86162.4 s = 2πr/v let this be equation 2

a) Determine the altitude of these satellites above the surface of the earth in both SI and U.S. customary units

we substitute v in equation into equation 2

so

86162.4 s = 2πr / (19.95 × 10⁶ / √r)

r = 42.16 × 10⁶ m

so, altitude h = r - R

h = 42.16 × 10⁶ m - 6.37 × 10⁶ m

h = 35790000 m

convert to kilometer

h = 35790000 / 1000

h = 35790 km

Convert to miles

h = 35790 / 1.609

h = 22243.63 miles

Therefore, the altitude of these satellites above the surface of the earth;

35790 km ( SI units )

22243.63 miles ( U.S. customary unit )

answer: 2×10^-2kg

explanation: density= mass per volume

vol=mass/density

converting to s.i units we have;

3÷1000kg = 0.003

1.5÷100m = 0.15

0.003/0.15 = 0.02kg