Physics Problem?????

Question

A Scientists want to place a 2600 kg satellite in orbit around Mars. They plan to have the satellite orbit a distance equal to 2.2 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-m^2/kg^2

 

This What I found and calculated

 

time it take the satellite to complete one revolution = 9.2 hours 

The speed the satellite have to be in a perfectly circular orbit = 1985 m/s

 

Well the question I just can't get the right answer for.

 

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 the correct way to solve it is by using Kepler's third law,

I arrange a equation like this

 

((T_1)²/(r₁)²)= ((T₂)²/(r₂)³)

 

But when I solve it,  I just keep getting the wrong answer, maybe I am plugging in the numbers wrong. 

 

 

Answer 1

In solving this problem, I used an approach similar to Soumendra. We

know centripetal force is  mV^2/r and gravitational attraction is

Gm*M/r^2, where m is the mass of the satellite and M is the mass of mars. Since a stable orbit requires the two opposing forces be equal, then mv^2 / r = Gm*M/r^2. Multiplying both sides by r, and dividing both sides by m gives us V^2 = GM/r.  then, V = sqrt(GM/r). Plugging

in for G M & r we get

V = sqrt( 6.67384 * 10 ^ -11 x 6.4191 *  10 ^ 23  / 10.8704 * 10 ^ 6)

 

When we sort out the powers of 10, this becomes:

 

V = sqrt(6.67384 x 6.4191 / 10.8704     * 10 ^ 6)

   = sqrt(3.941 * 10 ^ 6)

   = 1.9852 * 10 ^ 3  ,  or, 1985.2 meters / second

 

I calculated a slightly longer period, however.

The required circumference, or the orital circle is 2 * PI * 3.2 radius of mars.   3.2 radius of mars is 10.8704 * 10 ^ 6 => 10,870,400

multiply by 2 * PI => 68,300,738

 

When you divide this distance traveled by the calculated velocity, you

get an orbital period of 34,408 seconds, or 9.56 hours.

 

To calculate the velocity and distance, for an orbital period 8 times this large we can still use this equation, even though there are now two unknowns. This requires a technique we often use, in engineering, known as iteration. You choose values for one of the unknowns, in this case, r and see what the chosen value produces for an answer. We know that r must be greater than for the first part as centripetal force and circular velocity must both decrease, as r increases, since less centripetal force will be needed as the gravitational pull decreases, with increased distance. I solved this problem using about 6 iterations, and arrived at an answer very siimilar to what soumendra did, at about 12.78 times the radius of mars, from the center of the planet.

Answer 2

I don't know about Soumndra's answer, but actually I am not sure of the question.  To me the distance you asked for is simply as Soumendra said: 3.2rmars.  That's it, isn't it?

 

The radius should be about 1.087 x 10^7 meters, or 1.087e7 m.
I did not using all of your sig.figs.
Actually I am not sure of the question you are asking. Respectfully, before I put a lot of time into a problem, I am going to want to know exactly what is meant by 8 times longer - than what?

So my simple answer, arrived at as follows, is the first part of Soumendra's answer.

Radius of the orbit in space is the radius of mars + 2.2 times the radius of mars:
Rs = 3.397e6meters + 2.2 (3.397e6 meters)
= 1.087e7 meters.

But now I am curious. Let me see if this supports the other data that you gave. I think I am right, but just because I live here at the space coast means nothing. I’m new here, just got interviewed to work as a custodian for Walmarts, know no one here and I am on my own with this problem. Q:?)

I will work through several pieces of data that you listed, will work around Kepler's Law and I will be skipping many of the cumbersome dimensions.

Let me define the following variables and constants as follows so subscripts will be lower case:
Gu = Universal Gravitational constant = 6.67e-11
Mmars = mass of planet Mars = 6.42e23 kg
Rmars = radius Mars’ surface = 3.397e6 meters
Rs = radius of the satellite's orbit in space around Mars = 1.087e7 meters
Gmars = Gravity on the surface of mars which we need to calculate
Gs = Gravitational acceleration at a distance from Mars equal to the orbiting satellite’s altitude

You got around nine hours for the period of the orbit. Let me see what I can get using the answer 1.087e7.

The period of the orbit is circumference divided by the linear speed of the satellite in orbit.

The circumference of the orbit having radius 1.087e7 is C = 2 pi r:

C = 2 * pi * 1.087e7 meters = 6.83e7 meters.

Now we need the velocity and see if the circumference divided by the velocity is about nine hours.

To get the velocity we might use the formula v = sqr(rg), where r is the radius of the satellite’s orbit and g is Gs.

But first we need Gs. We get Gs from the proportion Gs/Gmars = Rmars^2/Rs^2 because gravity is … indirectly proportional to the square of the distance between centers….

And now note we need Gmars to get Gs!

We get Gmars from Gmars / Mmars = Gu / Rmars.

Gravity on mar's surface is Gmars = (Gu * Mmars) / Rmars.

Gravity on mar's surface is Gmars = (6.67e-11 * 6.42e23) / (3.397e6)^2 = 3.71 meters per second squared.

Now back to find gravity in space at the satellite’s orbital radius:

Gs / Gmars = Rmars^2 / Rs^2, again because gravity is … indirectly proportional to the square of the distance between centers….

Solving for Gs: Gs = Gmars * Rmars^2 / Rs^2

Subbing Gs = 3.71 * 3.397e6^2 / 1.087e7^2 for Gs:

Gs = 0.362 m/s^2;

Now we have the ingredients for the simple formula V = sqr( r g ).

V = linear instantaneous velocity of the satellite in orbit around Mars where r = 1.087e7 meters and g = 0.362 m/s^2:

V = sqr ( 1.087e7 * 0.362 ) m/s

V = 1984 m/s

This is the linear speed of the satellite at 1.087 m from the center of Mars.

Finally the time to complete one orbit is Circumference / velocity:

T = (6.83e7 / 1984) s

T = 34400 s

T = 9.6 hours.

You got about 9.2. I am in the ball park, but maybe not close enough.

Anyway, it might be worth going through. I have a feeling it might be good to review some of the basics I used and you might find an error – probably in my thinking.

Answer 3

For a satellite of mass m, orbiting at a distance r from the center centrpetal force = gravitationa pull:

GmM/r² =mw²r

 

 or GM =w²r³ = (2Pi/T)² * r³

 

or (T₁/T₂)² = (r1/r2)³

 

Given TT2/T1 = 8, r1 = 2.2rmars + rmars = 3.2rmars

(r2/r1)³ = 8 ² = 64 = 4³

(r2/r1) = 4

r2 = 4 * r1 = 4 *3.2rmars = 12.8rmars

 

So distance of  the satellite from the center of  Mars = 12.8rmars = 12.8 * 3.397 x 10⁶ m  = 4.348 x 10⁷ m