# Grade 12 Gravitational Force Fields Questions.

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

Related Introductory Physics Homework Help News on Phys.org
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.

gneill
Mentor
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.

gneill
Mentor
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.

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!

Thanks for the correction!
So what's going to be changed in the calculation process (how do you find the new velocity?)

gneill
Mentor
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.

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.

gneill
Mentor
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.

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.

gneill
Mentor
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.

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!

gneill
Mentor
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.