Answer:
(a)F= 3.83 * 10^3 N
(b)Altitude=8.20 * 10^5 m
Explanation:
On the launchpad weight = gravitational force between earth and satellite.
W = GMm/R²
where R is the earth radius.
Re-arranging:
WR² / GM = m
m = 4900 * (6.3 * 10^6)² / (6.67 * 10^-11 * 5.97 * 10^24) = 488 kg
The centripetal force (Fc) needed to keep the satellite moving in a circular orbit of radius (r) is:
Fc = mω²r
where ω is the angular velocity in radians/second. The satellite completes 1 revolution, which is 2π radians, in 1.667 hours.
ω = 2π / (1.667 * 60 * 60) = 1.05 * 10^-3 rad/s
When the satellite is in orbit at a distance (r) from the CENTRE of the earth, Fc is provided by the gravitational force between the earth and the satellite:
Fc = GMm/r²
mω²r = GMm / r²
ω²r = GM / r²
r³ = GM/ω² = (6.67 * 10^-11 * 5.97 * 10^24) / (1.05 * 10^-3)²
r³ = 3.612 * 10^20
r = 7.12 * 10^6 m
(a)
F = GMm/r²
F=(6.67 * 10^-11 * 5.97 * 10^24 * 488) / (7.12 * 10^6 )²
F= 3.83 * 10^3 N
(b) Altitude = r - R = (7.12 * 10^6) - (6.3 * 10^6) = 8.20 * 10^5 m
