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Scientists want to place a 3700 kg satellite in orbit around mars. they plan to have the satellite orbit a distance equal to 1.9 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?

Respuesta :

skyluke89
The force of attraction between Mars and the satellite is:
[tex]F=G \frac{Mm}{d^2} [/tex]
where G is the gravitational constant, M the mass of Mars, m the mass of the satellite, and d the distance of the satellite from the center of Mars.

First we have to calculate d. Calling r the radius of Mars, we know that the satellite is located at 1.9r above the surface. This means that its distance from the center of Mars is [tex]d=1.9r +r=2.9 r[/tex].
Using this information, and the data given by the problem, we can now calculate the intensity of the gravitational attraction:
[tex]F=G \frac{Mm}{(2.9r)^2}=(6.67 \cdot 10^{-11}Nm^{-2}kg^{-2}) \frac{(6.42\cdot 10^{23}kg)(3700kg)}{(2.9\cdot3.39 \cdot 10^6m)^2}=1639 N [/tex]

Answer:

3805.59 N

Explanation:

Parameters given:

Mass of satellite, m = 3700 kg

Mass of Mars, M = [tex]6.4191 * 10^{23} kg[/tex]

Radius of Mars, r = [tex]3.397 * 10^6[/tex] m

Distance between the satellite and the surface of Mars, D = 1.9 times r

D = [tex]1.9 * 3.397 * 10^6[/tex] = [tex]6.454 * 10^6 m[/tex]

The gravitational force of attraction is given as:

[tex]F = \frac{GMm}{D^2}[/tex]

where G = gravitational constant = [tex]6.67428 * 10^{-11}Nm^2/kg^2[/tex]

[tex]F = \frac{6.67428 * 10^{-11} * 6.4191 * 10^{23} *3700}{(6.454 * 10^6)^2}[/tex]

[tex]F = 3805.59 N[/tex]

The gravitational force of attraction is 3805.59 N

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