Scientists want to place a 4 x 10 kg satellite in orbit around Mars. They plan t

Question

Scientists want to place a 4 x 10 kg satellite in orbit around Mars. They plan to have the satellite orbit a distance equal to 2.4 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 10 kg rman 3.397 x 10 m G-6.67428 x 10 N-m2/kg 1) What is the force of attraction between Mars and the satellite? Submit You currently have 0 submissions for this question. Only 9 submission are allowed. You can make 9 more submissions for this question. 2) What speed should the satellite have to be in a perfectly circular orbit? m/s Submit You currently have O submissions for this question. Only 9 submission are allowed You can make 9 more submissions for this question 3) How much time does it take the satellite to complete one revolution? rs Submit You currently have O submissions for this question. Only 9 submission are allowed. You can make 9 more submissions for this question. 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 Submit You currently have O submissions for this question. Only 9 submission are allowed You can make 9 more submissions for this question. 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? Submit You currently have O submissions for this question. Only 9 submission are allowed. You can make 9 more submissions for this question.

Explanation / Answer

mass of the satellite, m1=4*10^3 kg

mass of the Mars, m2=6.4191*10^22 kg

radius of the circular path,r=2.4*r_mars

r=2.4*3.397*10^6 m

1)

gravitational force, Fg=G*m1*m2/r^2

F=6.67428*10^-11*(4*10^3)*(6.4191*10^22)/(2.4*3.397*10^6)^2

F=257.825 N

2)

use,

m1*v^2/r=Fg

4*10^3*V^2/(2.4*3.397*10^6)=257.825

===> v=724.913 m/sec

3)

time period, T=2pi*r/v

=2pi*(2.4*3.397*10^6)/(724.913)

=19.629 hr

4)

mass of the planet

and

radius of the orbit

5)

use,

r^3/T^2=consatnt,

==> (r2/r1)^3=(T2/T1)^2

===>(r2/(2.4*3.397*10^6))^3=(8*T1/T1)^2

===>(r2/(2.4*3.397*10^6))^3=(8)^2

==> r2=32.6112*10^6

new radius, r2=32.6112*10^6

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