Scientists want to place a 3100 kg satellite in orbit around Mars. They plan to

Question

Scientists want to place a 3100 kg satellite in orbit around Mars. They plan to have the satellite orbit at a speed of 2598 m/s in a perfectly circular orbit. Here is some information that may help solve this problem: m_mars = 6-4191 x 10^22 kg r_mars:i = 3397 X 106 m G = 6.67428 x 10^ 11 N-m2/kg2 What radius should the satellite move at in its orbit? (Measured from the center of Mars.) m Submit Your submissions: What is the force of attraction between Mars and the satellite? N | Submit | Your submissions:What is the force of attraction between Mars and the satellite? N | Submit | what is the force of attraction between Mars and the satellite? m/s^2 submit which of the following quantites would change the radius the satellite needs to orbit at? the mass of the satellite the mass of the planet the speed of the satellite what should the speed of the orbit be, if we want the satellite to take 8 times longer to complete one full revolution of its orbit? m/s Submit

Explanation / Answer

1)

The radius (r) =Gmmars/v2= (6.67428*10-11)(6.4191*1023)/(2598)2 =6.347*106m

2)

Now the force of attraction between the mars and satellite is

F =GMm/R2 =(6.67428*10-11)(6.4191*1023)(3100)/(3.397*106)2 =11509.306N

3)

The acceleration of the satellite in the orbit is

a =GM/R2=(6.67428*10-11)(6.4191*1023)/(3.397*106)2 =3.7126m/s2

4)

Now from the kepler's third law

T22/T12 = R23/R13

(8 T1/ T1)2 = (R2 / (3.397x 106 ))3

R2 = 9.02 x 106 m

New orbital speed is given as

V' = sqrt(Gmmars/R2)

V' = sqrt(6.67428 x 10-11 x 6.4191x 1023 / (9.02x 106 ))

V' = 2179 m/s

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