Scientists want to place a 2500 kg satellite in orbit around Mars. They plan to have the satellite orbit a distance equal to 2.1 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?

Respuesta :

Answer:

Part (i) the force of attraction between Mars and the satellite is 965.78 N

Part (ii)  the speed of the satellite in a perfectly circular orbit 2016.99 m/s

Part(iii)  time it takes the satellite to complete one revolution is 32809.78 s

Part (iv) the radius of the orbit

Part (v) the radius of the orbit is 8.425 x 10⁷ m

Explanation:

Part (i) the force of attraction between Mars and the satellite

Given;

mass of Mars, m₁=  6.4191 x 10²³ kg

mass of satellite, m₂ = 2500 kg

Distance between the satellite and Mars, R = radius of mass + 2.1 times the radius of Mars = 3.397 x 10⁶ m + 2.1(3.397 x 10⁶ m) = 1.0531 X 10⁷ m

[tex]F = \frac{G*M_1M_2}{R^2}[/tex]

Where;

G is gravitational constant =  6.67428 x 10⁻¹¹ Nm²/kg²

[tex]F = \frac{G*M_1M_2}{R^2} = \frac{6.67428 X 10^{-11}*(6.4191 X 10^{23})(2500)}{( 1.0531 X 10^7)^2}\\\\F = 965.78 N[/tex]

Part (ii)  the speed of the satellite in a perfectly circular orbit

According to Newton's second law;

F = ma  and a = v²/R

[tex]M_2\frac{V^2}{R} = \frac{G*M_1M_2}{R^2} \\\\V^2 = \frac{G*M_1}{R} \\\\V= \sqrt{\frac{GM_1}{R}} =\sqrt{\frac{ (6.67428 X 10^{-11})(6.4191 X10^{23})}{1.0531 X10^7}} \\\\V = \sqrt{4068262.3443} = 2016.99 \frac{m}{s}[/tex]

Part(iii)  time it takes the satellite to complete one revolution

one complete revolution = 2πR

                                          = 2π X 1.0531 X 10⁷ m = 6.6177 X 10⁷ m

[tex]T = \frac{D}{V} =\frac{6.6177 X 10^7}{2016.99} = 32809.78 s[/tex]

Part (iv)

From the formula above, it is the mass of Mars and Radius of orbit.

Thus, the correct answer is the radius of the orbit

Part (v)

If it takes 8 times longer, T = 8 X32809.78 s = 262478.24 s

D = T*V

   = 262478.24* 2016.99 = 529415985.3 m

D = 2πR

R = D/2π

   = (529415985.3)/2π

   = 8.425 x 10⁷ m

Mars and the satellite is 965.78 N, perfectly circular orbit 2016.99 m/s. complete one revolution is 32809.78 s, the radius of the orbit, the radius of the orbit is 8.425 x 10⁷ m.

What is the satellite orbit ?

The orbit of the satellite is the high and low orbits that can be taken as polar and equatorial orbits. It consists of the spin axis of the earth. The orbits are then planes that coincide with the direction of the satellite motion. The satellite of 2500 kg is around mars.

Part (i) the force of the attraction between the Mars and the satellite

mass of Mars, m₁=  6.4191 x 10²³ kg and the mass of satellite, m₂ = 2500 kg. Distance between the satellite and Mars, R = radius of the mass + 2.1 times the radius of Mars equal to  3.397 x 10⁶ m + 2.1(3.397 x 10⁶ m) = 1.0531 X 10⁷ m Here,  the G is the gravitational constant is  6.67428 x 10⁻¹¹ Nm²/kg²

Part (ii)  the speed of the satellite in the perfectly circular orbit

As per the Newton's 2nd law; F = ma  and a = v²/R

Part(iii)  time it takes the satellite to completes 1 revolution

one complete revolution   = 2π X 1.0531 X 10⁷ m = 6.6177 X 10⁷ m

Part (iv)

From above, it is the M of Mars and the Radius of orbit.

Hence , the answer is in the radius of the orbits.

Part (v)

About 8 times longer, T is 8 X32809.78 s = 262478.24 s

D is T*V   = 262478.24* 2016.99 = 529415985.3 m

D equals 2πR  R = D/2π   = (529415985.3)/2π

Hence    = 8.425 x 10⁷ m

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