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western50 · Jan 29, 2011, 08:01 PM

An 140000 kg jet accelerates down the runway for takeoff at 2.1 m/s^2. Once off the ground, the plane climbs upward for 20 seconds. During this time, the vertical speed increases from zero to 17 m/s, while the horizontal speed increases from 80 m/s to 95 m/s. After reaching cruising altitude, the plane levels off, keeping the horizontal speed constant, but smoothly reducing the vertical speed to zero, in 11 seconds.

What is the net horizontal force on the airplane as it levels off? 0 Newton
What is the net vertical force on the airplane as it levels off? -216363.64 Newton
Am I right?

Scientists want to place a 3900 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:

m(mars) = 6.4191 x 1023 kg
r(mars) = 3.397 x 106 m
G = 6.67428 x 10-11 N-m2/kg2 and it take 9.11 hours to make one complete revolution.

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?

I think it should be 5.29*10^8m, am I right?

jcaron2 · Jan 29, 2011, 10:05 PM

For question 1, I agree with both of your answers.

For question 2, if I remember correctly back to my days of astrophysics in college, Kepler's third law states that the orbital period squared is proportional to the orbital radius cubed (for circular orbits, where orbital radius is measured from the center of the planet to the satellite):

$T^2 \propto r^3$

Therefore you can write the equation

$\frac{(T_1)^2}{(r_1)^3} = \frac{(T_2)^2}{(r_2)^3}$ ,

where $T_1$ is the period given (9.11 hours, but it doesn't matter because that number should cancel out from both sides of the equation) and $r_1$ is the radius of the given orbit from the center of the planet (which makes a total of 3.1*r_mars)

Then it's just a matter of plugging in $T_2 = 8 \cdot T_1$ , and solving for $r_2$ .

Let us know what you got for an answer.

Unknown008 · Jan 30, 2011, 07:09 AM

Well, if you really have to use the values given to you...

$F = \frac{GMm}{r^2}$

And

$F = mr\omega^2$

will be helpful.

Though I agree that working all this out is time consuming, you can then derive Kepler's law.

$\frac{GMm}{r^2} = mr\omega^2$

$\frac{GM}{r^3} = \omega^2$

$\frac{4\pi^2}{T^2} = \frac{GM}{r^3}$

So, basically,

$r^3 \propto T^2$

since G, M, 4 and pi are constants.