www.flipitphysics.com/ Scientists want to place a 3 x 103 kg satellite in orbit

Question

www.flipitphysics.com/ Scientists want to place a 3 x 103 kg satellite in orbit around Mars. They plan to have the satellite orbit a distance equal to 1.8 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 mars 3.397 x 106 m G 6.67428 x 10-11 N-m2/kg2 What is the force of attraction between Mars and the satellite? What speed should the satellite have to be in a perfectly circular orbit? How much time does it take the satellite to complete one revolution? oWhich of the following quantities would change the speed the satellite needs to orbit at? 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

1) f =Gm1m2/r^2

f = 6.67*106-11*3*10^3*6.4191*10^23/(1.8*3.397*10^6)^2

f = 3.435*10^3 N

b) v = sqrt(Gm/r) = sqrt(6.67*10^-11*6.4191*10^23/1.8*3.397*10^6) = 2.65*10^3 m/s

c) v = 2pi*r/T

T = 2*3.14*1.8*3.397*10^6/2.65*10^3 = 8050.2 sec = 2.24 hrs

d) v is mainly depends onthe planet mass and radisu of the orbit only

e) v1/v2 = r1T2/r2T1

r2 = r1*T2*v2/v1*T1

V1/V2 = sqrt(r2/r1)

r2 = v1^2*r1/v2^2

first find v2 and then we can find r2

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