Gravitational pull on asteroid/Period it takes for one orbit

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  • #1
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Homework Statement


A bit of asteroid material of mass 2.5 x 10^6 kg orbits around the earth at an altitude of 2800km. How long does it take the asteroid to complete one orbit.

M asteroid = 2.5 x 10^6 kg
altitude of asteroid = 2800 km
radius asteroid = 200m
radius earth = 6.37 x 10^6 m


Homework Equations


Fnet = mv^2/r
C = 2∏r
t = d/v



The Attempt at a Solution


r = rEarth + rAsteroid + altitude
r = 9.2 x 10^6 m

C = 2∏r
C = 5.78 x 10^7
C = d

Fg = mv^2/r
mg = mv^2/r
9.8 = v^2/r
9.8 (9.2 x 10^6) = v^2
v = 9500 m/s

t = d / v
t = 5.78 x 10^7 / 9500
t = 6084 s
t = 1.7 h

My friend got 0.29 and the teacher had told him he was right. I didn't have a chance to see what I did wrong so I'm wondering if anyone here can tell me where I made my mistake.
 

Answers and Replies

  • #2
gneill
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Use Newton's law of gravitation to find the gravitational force when the object is not very close the the surface of the Earth; f = m*g just won't cut it here.
 
  • #3
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That would be

F = G * M * m/r if I'm not mistaken?

Would F be mg still? Also how does that help me find the velocity?
 
  • #4
gneill
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That would be

F = G * M * m/r if I'm not mistaken?

Would F be mg still? Also how does that help me find the velocity?

The r should be squared in your formula.

No, F is not mg. You just wrote the equation for the force due to gravity. Use that where you used mg in your work.
 
  • #5
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I think I know what you mean, would it be like this.

mv^2/r = (G)(M)(m)/r^2

v^2/r = (G)(M)/ r^2

v^2 = (G)(M)/r

Before I substitute values in, how do I know what to put for M and what m would be? I know that the two m would cancel out and that was the mass of the asteroid, so then M should be the mass of the earth. Generally speaking though, how do I know what mass to use?

Edit:

I tried substituting values in and I got 2.4 h which is still wrong apparently.

Work:

v^2 = GM/r
v^2 = (6.67 x 10^-11)(5.96 x 10^24) / 9.2 x 10^6
V = 6573

t = d/v
t = 5.78 x 10^7 / 6573
t = 8794 s
t = 2.4 h
 
Last edited:
  • #6
gneill
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20,925
2,866
I think I know what you mean, would it be like this.

mv^2/r = (G)(M)(m)/r^2

v^2/r = (G)(M)/ r^2

v^2 = (G)(M)/r

Before I substitute values in, how do I know what to put for M and what m would be? I know that the two m would cancel out and that was the mass of the asteroid, so then M should be the mass of the earth. Generally speaking though, how do I know what mass to use?

That looks better. The small m, which was also used in the expression for the centripetal force, represents the mass of the orbiting body. The large M is the mass of the primary, in this case the mass of the Earth.
 
  • #7
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Ah, I editted the other post not thinking you had seen my previous post. Anyways..

I tried substituting values in and I got 2.4 h which is still wrong according to what I was told.

Work:

v^2 = GM/r
v^2 = (6.67 x 10^-11)(5.96 x 10^24) / 9.2 x 10^6
V = 6573

t = d/v
t = 5.78 x 10^7 / 6573
t = 8794 s
t = 2.4 h
 
  • #8
gneill
Mentor
20,925
2,866
Ah, I editted the other post not thinking you had seen my previous post. Anyways..

I tried substituting values in and I got 2.4 h which is still wrong according to what I was told.

Work:

v^2 = GM/r
v^2 = (6.67 x 10^-11)(5.96 x 10^24) / 9.2 x 10^6
V = 6573

t = d/v
t = 5.78 x 10^7 / 6573
t = 8794 s
t = 2.4 h

Your answer looks fine to me.
 

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