Grade 12 Gravitational Force Fields Questions.

  • Thread starter TheSerpent
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  • #1
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1.Nasa fires a 1 tonne space probe, from a stable orbit of altitude 13,620km using a giant rail gun. The rail gun provides the probe with 9.97165*10^10 J of additional kinetic energy. Determine if the probe has enough velocity to escape the earth's gravity well. The Earth's mass and radius are 5.98*10^24kg and 6380km respectively.
 

Answers and Replies

  • #2
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Hi TheSerpent,
The formula for escape velocity is v=√(2GM/r) where v= velocity, G=universal gravitational constant (G=6.67*10-11 Nm2kg-2), M is the mass of the body, and r is the separation distance.

v=√(2GM/r)
Ekinetic=0.5mv2

m (rocket)=1000 kg
r (orbit)=13620 km
Ekinetic=9.97165*1010 J
m (earth)=5.98*1024 kg
r (earth)=6380 km

First, find the velocity that the rocket has...
Ekinetic=0.5mv2
9.97165*1010 J = 0.5(1000 kg)v2
v=√((2*(9.97165*1010 J))/1000 kg)
v=1.412*104 m/s

Then, find the velocity that the rocket would need to escape...
v=√(2GM/r)
v=√(2GM/rorbit+rearth)
v=√((2*(6.67*10-11 Nm2kg-2)*(5.98*1024 kg))/(13620 km + 6380 km))
v=1.997*105 m/s

Compare the two -- the rocket will not have sufficient velocity to escape.
Good luck! I hope this helps.
 
  • #3
gneill
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First determine the initial velocity of the space probe in its original orbit (prior to receiving a boost from the rail gun). Remember that you're given its altitude, not its orbit radius.

From that velocity determine what the escape velocity would be; escape velocity has a particular relationship to the circular orbit velocity. Then determine what the new velocity would be given the rail gun boost of energy. Compare to the escape velocity.
 
  • #4
gneill
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Hi TheSerpent,
The formula for escape velocity is v=√(2GM/r) where v= velocity, G=universal gravitational constant (G=6.67*10-11 Nm2kg-2), M is the mass of the body, and r is the separation distance.

v=√(2GM/r)
Ekinetic=0.5mv2

m (rocket)=1000 kg
r (orbit)=13620 km <--- That's its altitude, not its orbital radius
Ekinetic=9.97165*1010 J <--- That's the energy added, not the total
m (earth)=5.98*1024 kg
r (earth)=6380 km

First, find the velocity that the rocket has...
Ekinetic=0.5mv2
9.97165*1010 J = 0.5(1000 kg)v2
v=√((2*(9.97165*1010 J))/1000 kg)
v=1.412*104 m/s

Then, find the velocity that the rocket would need to escape...
v=√(2GM/r)
v=√(2GM/rorbit+rearth)
v=√((2*(6.67*10-11 Nm2kg-2)*(5.98*1024 kg))/(13620 km + 6380 km))
v=1.997*105 m/s

Compare the two -- the rocket will not have sufficient velocity to escape.
Good luck! I hope this helps.
The rocket begins in a circular orbit at the given altitude. Therefore it will have an initial velocity and kinetic energy appropriate for a circular orbit at that location. Next additional energy is supplied via a rail gun boost. The probe will gain that energy as KE and thus have a new velocity. Compare to the escape velocity for that orbital radius.
 
  • #5
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The rocket begins in a circular orbit at the given altitude. Therefore it will have an initial velocity and kinetic energy appropriate for a circular orbit at that location. Next additional energy is supplied via a rail gun boost. The probe will gain that energy as KE and thus have a new velocity. Compare to the escape velocity for that orbital radius.
Thanks for the correction!
 
  • #6
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Thanks for the correction!
So what's going to be changed in the calculation process (how do you find the new velocity?)
 
  • #7
gneill
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So what's going to be changed in the calculation process (how do you find the new velocity?)
The rail gun adds kinetic energy. Velocity depends upon kinetic energy.
 
  • #8
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The rocket begins in a circular orbit at the given altitude. Therefore it will have an initial velocity and kinetic energy appropriate for a circular orbit at that location. Next additional energy is supplied via a rail gun boost. The probe will gain that energy as KE and thus have a new velocity. Compare to the escape velocity for that orbital radius.
Find orbital velocity:
v = √(GM/r)
v = √((6.67*10^-11)(5.98*10^24)/(13620000 + 6380000)
v = 4.466 * 10^3 m/s

Find orbital kinetic energy:
K = mv^2/2
k = (1000)(4.466*10^3)^2/2
K = 9972578000 = 9.973 * 10^9 J

Find escape velocity:
v esc = √2 v
v esc = √2 (4.466 * 10^3)
v esc = 6.316 * 10^3 m/s

How would you include the energy boost from the rail gun into the answer, that is what I am not sure of.
 
  • #9
gneill
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Find orbital velocity:
v = √(GM/r)
v = √((6.67*10^-11)(5.98*10^24)/(13620000 + 6380000)
v = 4.466 * 10^3 m/s

Find orbital kinetic energy:
K = mv^2/2
k = (1000)(4.466*10^3)^2/2
K = 9972578000 = 9.973 * 10^9 J

Find escape velocity:
v esc = √2 v
v esc = √2 (4.466 * 10^3)
v esc = 6.316 * 10^3 m/s
Okay! Good to here!
How would you include the energy boost from the rail gun into the answer, that is what I am not sure of.
Kinetic energy adds. Add the boost energy to the energy k that you found above. Find the new velocity from that.
 
  • #10
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Okay! Good to here!


Kinetic energy adds. Add the boost energy to the energy k that you found above. Find the new velocity from that.
Find new velocity:

K = mv^2 / 2
(9.973 * 10^9)+(9.97165 * 10^10) = (1000)v^2 / 2
v = √(2)(1.096895*10^11)/1000
v = 14811.44827 = 1.481 * 10^4 m/s

This is the actual velocity so therefore the rocket will have sufficient velocity to escape.
 
  • #11
gneill
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Find new velocity:

K = mv^2 / 2
(9.973 * 10^9)+(9.97165 * 10^10) = (1000)v^2 / 2
v = √(2)(1.096895*10^11)/1000
v = 14811.44827 = 1.481 * 10^4 m/s

This is the actual velocity so therefore the rocket will have sufficient velocity to escape.
Yup. That looks good.
 
  • #12
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Yup. That looks good.
Thank you for your help!!!! Can you also help us in our other post you commented on? We posted an idea but unsure!
 
  • #13
gneill
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Thank you for your help!!!! Can you also help us in our other post you commented on? We posted an idea but unsure!
You're welcome.

I see that you have two threads, this one and one other. I've made a suggestion for you to work on in that other thread.
 

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