Scientists want to place a 3900 kg satellite in orbit around Mars. They plan to
ID: 1431344 • Letter: S
Question
Scientists want to place a 3900 kg satellite in orbit around Mars. They plan to have the satellite orbit a distance equal to 2.2 times the radius of Mars above the surface of the planet. Here is some information that will help solve this problem:
mmars = 6.4191 x 1023 kg
rmars = 3.397 x 106 m
G = 6.67428 x 10-11 N-m2/kg2
1) What is the force of attraction between Mars and the satellite?
2) What speed should the satellite have to be in a perfectly circular orbit?
3) How much time does it take the satellite to complete one revolution?
4) Which of the following quantities would change the speed the satellite needs to orbit at?
the mass of the satellite?
the mass of the planet?
the radius of the orbit?
5) What should the radius of the orbit be (measured from the center of Mars), if we want the satellite to take 8 times longer to complete one full revolution of its orbit?
Explanation / Answer
here,
mass of satellite , m = 3900 Kg
Mm = 6.419 *10^23 Kg
r = 3.397 *10^6 m
1)
force of attraction between satellite and mars = G *m1 * m2/d^2
force of attraction between satellite and mars = 6.673 *10^-11 * 6.419 *10^23* 3900/(2.2 * 3.397 *10^6 )^2
force of attraction between satellite and mars = 2991 N
the force of attraction between satellite and mars is 2991 N
2)
let the speed of satellite is v
m * v^2/r = force of attraction
3900 * v^2/(2.2 * 3.397 *10^6) = 2991
solving for v
v = 2394 m/s
the speed of satellite is 2394 m/s
3)
time taken for one revolution , t = 2*pi * r/v
time taken for one revolution , t = 2 * pi * 2.2 * 3.397 *10^6/2394
time taken for one revolution , t = 19610 s
the time taken for one revolution is 19610 s
4)
for speed
G* M * m r^2 = m* v^2/r
v = sqrt(G*M/r)
the speed of satellite dependent at
the mass of the planet
the radius of the orbit
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