Calculating Speed of MRO Orbiting Mars

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In summary, the conversation discusses a question about the speed of the Mars Reconnaissance Orbiter (MRO) in its orbit around Mars. The relevant equations and values are provided, including the mass of Mars, the mass of MRO, the distance between the two, the radius of Mars, and the gravitational constant. The correct answer is 3.5km/s, but the individual is getting a different answer of 27km/s. After some discussion and clarification, it is determined that the individual is not considering the units correctly, and once they make the necessary conversions, they are able to get the correct answer.
  • #1
462chevelle
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Homework Statement



what is the speed of MRO in its orbit around mars.
M mars=6.4X10^23
m MRO=907 Kg
d between the two=200 km
R of mars=3400 km
G=6.7X10^-11

Homework Equations


D=(mMG)/r^2=(mV^2)/R=(MG/R)^1/2



The Attempt at a Solution


I attached a pic of some of my work. if I need to add more detail or its hard to read let me know. I am just curious where the mistake is.
thanks
Lonnie
 

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  • #2
to clarify since I noticed its hard to interpret. the answer I got was appx 27 km/s and the correct answer is 3.5km/s
thanks
Lonnie
 
  • #3
462chevelle said:
to clarify since I noticed its hard to interpret. the answer I got was appx 27 km/s and the correct answer is 3.5km/s
thanks
Lonnie

You are certainly right that I can't interpret that. Hence I have no idea how you got 27km/s instead of the correct answer. Can you explain clearly how you got 27km/s?
 
  • #4
if I use D=mV^2/R
I get 200km = (907(V^2))/3400
multiply 3400*200 then divide by 907
to get 749=V^2
then (749)^1/2=appx. 27
that's the only equation from lecture that has velocity in it is the reason I am using it. I would think I would need to use an equation with G or M in it. but I don't have one written down here
 
  • #5
462chevelle said:
if I use D=mV^2/R
I get 200km = (907(V^2))/3400
multiply 3400*200 then divide by 907
to get 749=V^2
then (749)^1/2=appx. 27
that's the only equation from lecture that has velocity in it is the reason I am using it. I would think I would need to use an equation with G or M in it. but I don't have one written down here

You don't understand what your equations mean. mV^2/R is the centripetal force exerted on the satellite by its rotation around the planet. It's not equal the distance from the planet. R is the distance to the center of the planet from the satellite, I hope you know what the rest mean. Now GMm/R^2 is the gravitational force. If the satellite is in a stable circular orbit then the two forces are equal. Set them equal and solve for V.
 
Last edited:
  • #6
youre correct. I wasn't understanding what I was doing. I guess I just started plugging stuff out of frustration.
I started over and set them equal to each other like so
mV^2/R=GMm/R^2
factored out useless variables and rewrote as
V=(GM/r)^1/2
solved It again and got crap numbers again. I think I need to get some sleep and try it tomorrow with a clear mind. I know I am doing something wrong that is very simple. I have to be. because in order to get 3.5 as velocity I will have to get somewhere around 12 as my answer inside of the parenthesis and that seems impossible with my lack of sleep or focus right now.
 
  • #7
462chevelle said:
youre correct. I wasn't understanding what I was doing. I guess I just started plugging stuff out of frustration.
I started over and set them equal to each other like so
mV^2/R=GMm/R^2
factored out useless variables and rewrote as
V=(GM/r)^1/2
solved It again and got crap numbers again. I think I need to get some sleep and try it tomorrow with a clear mind. I know I am doing something wrong that is very simple. I have to be. because in order to get 3.5 as velocity I will have to get somewhere around 12 as my answer inside of the parenthesis and that seems impossible with my lack of sleep or focus right now.

Yes, try it again. Now you've got the right formula, but keep units in mind. G is probably given in units of meters, put everything in the same units and expect to get ~3500 m/s as your answer.
 
  • #8
ahh crap.. i think you're right. i always screw up the units. after class ill have another stab at it and let you know how it goes.
thanks
 
  • #9
the only way i can get it to work out right is if i convert
6.7X10^-11 to 6.7X10^-20
and i figured that out by leaving G as a variable and plugging 3.5 into v and solving for G
i wouldn't think that i would have to move my units as far as 9 places. but as many values are in the constant g its hard for me to interpret how to convert it
my lesson says G=6.7X10^-11m^3kg^-1s^-2
i guess i should understand what all of those units are interpreted as.
 
  • #10
462chevelle said:
the only way i can get it to work out right is if i convert
6.7X10^-11 to 6.7X10^-20
and i figured that out by leaving G as a variable and plugging 3.5 into v and solving for G
i wouldn't think that i would have to move my units as far as 9 places. but as many values are in the constant g its hard for me to interpret how to convert it
my lesson says G=6.7X10^-11m^3kg^-1s^-2
i guess i should understand what all of those units are interpreted as.

The units on G and M are fine! kg,m,s. The only ones that aren't fine are 200km and 3400km. Just change those to meters.
 
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  • #11
awesome. that worked out perfect. thanks man. i would have to say units is something that messes me up a lot.
 

Related to Calculating Speed of MRO Orbiting Mars

1. How is the speed of the Mars Reconnaissance Orbiter (MRO) calculated?

The speed of the MRO is calculated using the equation v = √(GM/r), where v is the velocity, G is the gravitational constant, M is the mass of Mars, and r is the distance between the MRO and the center of Mars. This equation is derived from Newton's law of universal gravitation.

2. What is the current speed of the MRO orbiting Mars?

The MRO is currently orbiting Mars at an average speed of 3.8 kilometers per second. This speed may vary slightly depending on its distance from Mars and any maneuvers it may perform for scientific observations or to maintain its orbit.

3. How does the MRO maintain its speed while orbiting Mars?

The MRO maintains its speed by constantly falling towards Mars due to the planet's gravitational pull. As it falls, it also moves forward at a high velocity, resulting in a circular orbit around Mars. The MRO's thrusters are used to make small adjustments to its speed and trajectory to maintain its orbit.

4. Can the speed of the MRO be altered?

Yes, the speed of the MRO can be altered by using its thrusters to perform maneuvers. These maneuvers can change the MRO's speed and trajectory, allowing it to enter a different orbit or perform targeted observations of specific areas on Mars.

5. How does the speed of the MRO affect its scientific observations?

The speed of the MRO is crucial for its scientific observations. A faster speed allows the MRO to cover more ground and collect data from different areas of Mars. However, a slower speed may be necessary for more detailed observations and high-resolution images. The MRO's speed is carefully managed to balance these factors and maximize its scientific capabilities.

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