Answer to Question #141079 in Physics for Ann

Question

Two satellites A and B are orbiting the earth around the equator, E, at different altitudes on a circular path of 2,400 and 3,000 km from the surface of the earth. Take the radius of the earth ate the equator is Its equatorial radius is 6,378 km, mass of earth is 5.98*1024 kg, g=9.81m/s2, G= 6.67*10^-11 N m² /kg².N m² /kg². If the satellites are of the same mass, 100 Kg, determine the:

a) Force required to keep each satellites in orbit.

Answer 1

The force on the first sattelite is:

$F = G\dfrac{mM}{r_1^2}$

where $r_1 = 6.378km + 2400km = 8.778\times 10^6m$, $m= 100 kg$ is the mass of the sattelite and $M = 5.98\times 10^{24} kg$ is the mass of the Earth. Thus, obtain:

$F_1 = G\dfrac{mM}{r_1^2} = 6.67\times 10^{-11}\dfrac{100\cdot 5.98\times 10^{24}}{8.778^2\times (10^6)^2} \approx 517.65\space N$

Similarly, for the second sattelite $r_2 = 6.378km + 3000km = 9.378\times 10^6m$ and have:

$F_2 = G\dfrac{mM}{r_2^2} = 6.67\times 10^{-11}\dfrac{100\cdot 5.98\times 10^{24}}{9.378^2\times (10^6)^2} \approx 453.53\space N$

Answer. 517.65 N and 453.53 N.