• Support PF! Buy your school textbooks, materials and every day products Here!

Orbital speed

  • #1
462chevelle
Gold Member
305
9

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. im just curious where the mistake is.
thanks
Lonnie
 

Attachments

Answers and Replies

  • #2
462chevelle
Gold Member
305
9
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
Dick
Science Advisor
Homework Helper
26,258
618
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
462chevelle
Gold Member
305
9
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 im 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
Dick
Science Advisor
Homework Helper
26,258
618
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 im 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
462chevelle
Gold Member
305
9
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 im 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
Dick
Science Advisor
Homework Helper
26,258
618
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 im 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
462chevelle
Gold Member
305
9
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
462chevelle
Gold Member
305
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 wouldnt 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
Dick
Science Advisor
Homework Helper
26,258
618
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 wouldnt 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.
 
  • Like
Likes 1 person
  • #11
462chevelle
Gold Member
305
9
awesome. that worked out perfect. thanks man. i would have to say units is something that messes me up a lot.
 

Related Threads on Orbital speed

  • Last Post
Replies
1
Views
802
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
2
Views
7K
  • Last Post
Replies
1
Views
3K
  • Last Post
Replies
1
Views
6K
  • Last Post
Replies
2
Views
5K
  • Last Post
Replies
2
Views
758
Replies
1
Views
344
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
3
Views
893
Top