Is the Time of Orbit for a Satellite Above Mars Really 800 Million Years?

[Xanthippus]

Homework Statement

We have been asked to work out the approximate time of orbit, T, of a satellite orbiting above the centre of Mars, radius 4000km.

Homework Equations

We have been given the equation:

R^3/T^2 =1x〖10〗^13 m^3 s^(-2)

The Attempt at a Solution

When using the equation:

T=√((4x〖10〗^6 )^3/〖1x10〗^13 )

I get an answer of 2.53 x 10^16 seconds and I don't understand how something that seems to be accelerating so fast turns out to be taking 800 million years to orbit mars. Is my math just totally wrong?

EDIT: That maths was wrong I'm pretty sure, I now have an answer of 2529.8 seconds which is pretty fast for an orbit of Mars but 1x10^13 m^3 s^-2 is a hell of an acceleration.

 

[tiny-tim]

Hi Xanthippus!

EDIT: That maths was wrong I'm pretty sure, I now have an answer of 2529.8 seconds …

I think you're √10 out

 

[Xanthippus]

I don't see where...

T=√((〖4x10〗^6 )^3/〖1x10〗^13 )

T=√(〖64x10〗^18/〖1x10〗^13 )

T=√(〖64x10〗^5 )

T=2529.8s

 

[tiny-tim]

oops!

sorry, you're right!

 

[rwisz]

I would like to point out that the equation given in the homework is not the correct one for calculating the time of orbit of a satellite above Mars. The equation given, R^3/T^2 =1x〖10〗^13 m^3 s^(-2), is actually the third Kepler's law for planetary motion, which relates the orbital period (T) of a planet or satellite to its distance from the central body (R).

The correct equation for calculating the time of orbit for a satellite above Mars would be T=2π√(a^3/GM), where a is the semi-major axis of the satellite's orbit, G is the gravitational constant, and M is the mass of Mars.

Using this equation, we can calculate the time of orbit for a satellite at a distance of 4000km above the center of Mars. Assuming a circular orbit, the semi-major axis (a) would be equal to the radius of Mars plus the altitude of the satellite, so a=4000km + 3390km = 7390km.

Plugging in these values, we get T=2π√((7390km)^3/(6.67x10^-11 m^3/kg/s^2)(6.39x10^23 kg)) = 5543 seconds, or approximately 1.54 hours.

This is a much more reasonable time frame for the orbit of a satellite above Mars, compared to the 800 million years mentioned in the original content. Therefore, it is safe to say that the time of orbit for a satellite above Mars is not 800 million years, and the math used in the attempt at a solution was incorrect.

Furthermore, it is important to note that the equation given in the homework, R^3/T^2 =1x〖10〗^13 m^3 s^(-2), is only applicable for planetary motion, where the central body is much more massive than the orbiting body. In the case of a satellite orbiting Mars, the mass of the satellite is not negligible compared to the mass of Mars, so the equation cannot be used. This highlights the importance of using the correct equations and understanding the underlying principles when solving scientific problems.