# Grade 12 Force Field question

1. The meteoroid Perturbation passes near the planet Conniption which has a mass of 5 * 10^25kg, a charge of +10^6 C and has a radius of 10500km. The meteoroid has a mass of 10 tonnes and a charge of +5 C. The meteoriod has an altitude of 9500km and a velocity of 10km/s. The planets magnetic field is 0.001T perpendicular to the meteoroids velocity. Determine the instantaneous acceleration of the meteoroid.

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gneill
Mentor
Relevant Equations? Attempt at Solution?

Unsure on where to begin!

gneill
Mentor
Unsure on where to begin!
Begin with relevant equations. What forces are involved and from what fields?

The forces that are involved in this question from what I can tell are gravitational and magnetic.

Some possible relevant formulas are:
Fg = GMm / r^2
Fe = kQq / r^2
Fb = qvBSin(theta)

I think it would be required to substitute these into F = ma to solve for 'a' somehow, but I am not sure.

gneill
Mentor
The forces that are involved in this question from what I can tell are gravitational and magnetic.

Some possible relevant formulas are:
Fg = GMm / r^2
Fe = kQq / r^2
Fb = qvBSin(theta)

I think it would be required to substitute these into F = ma to solve for 'a' somehow, but I am not sure.
Okay, that's progress.

Each of those forces will have a contribution to the total force on the mass. Each force will act in a direction that depends upon the nature of the field. That is, each is a vector quantity and you will add the vectors to determine the net force. The acceleration can then be found via $\vec{a} = \vec{F}/m$

mPerturbation = 10 tonnes = 10000 kg
mConniption = 5 * 10^25 kg
charge of Conniption [Q] = +10^6 C
charge of Perturbation [q] = +5 C
rConniption = 10500000 m
altitude Perturbation = 9500000 m
r = rCon + altitPert = 20000000
v Perturbation = 10000 m/s
B Conniption = 0.001 T (perpendicular to meteoroid's velocity)
a = ?

Attempt at solution.

Find instantaneous acceleration:

F = ma
a = F / m
a = ( -Fg + Fe + Fb ) / m
a = [(-GMm/r^2) + (KQq/r^2) + (qvBsin90)]/m
a = [(-(6.67*10^-11)(5*10^25 kg)(10000 kg)/(20000000 m)^2) + ((9.0*10^9)(+10^6 C)(+5 C)/(20000000 m)^2) + ((+5 C)(10000 m/s)(0.001 T)(1))/(10000 kg)]
a = -8.32125 N/kg

Not sure if the vectors are absolutely correct though or if something is out of place.

gneill
Mentor
mPerturbation = 10 tonnes = 10000 kg
mConniption = 5 * 10^25 kg
charge of Conniption [Q] = +10^6 C
charge of Perturbation [q] = +5 C
rConniption = 10500000 m
altitude Perturbation = 9500000 m
r = rCon + altitPert = 20000000
v Perturbation = 10000 m/s
B Conniption = 0.001 T (perpendicular to meteoroid's velocity)
a = ?

Attempt at solution.

Find instantaneous acceleration:

F = ma
a = F / m
a = ( -Fg + Fe + Fb ) / m
a = [(-GMm/r^2) + (KQq/r^2) + (qvBsin90)]/m
a = [(-(6.67*10^-11)(5*10^25 kg)(10000 kg)/(20000000 m)^2) + ((9.0*10^9)(+10^6 C)(+5 C)/(20000000 m)^2) + ((+5 C)(10000 m/s)(0.001 T)(1))/(10000 kg)]
a = -8.32125 N/kg

Not sure if the vectors are absolutely correct though or if something is out of place.
You're answer looks okay for the given values, although the problem makes the gravitational force so much larger than the others that it swamps their effects.

I also note that the specification of the magnetic field is ambiguous -- it is directed perpendicular to the velocity of the meteoroid, but that only defines a plane in which the field lines lie, not the direction of the field within that plane. That makes a difference to the direction of the force that results; if the resultant magnetic force did not align with the others you'd have to add their vector components rather than simply sum their magnitudes. Even aligned the magnetic force could be negative rather than positive.

Forgot that Fg = -GMm / r^2 so that the - cancel out:

F = ma
a = F / m
a = ( -Fg + Fe + Fb ) / m
a = [-(-GMm/r^2) + (KQq/r^2) + (qvBsin90)]/m
a = [-(-(6.67*10^-11)(5*10^25 kg)(10000 kg)/(20000000 m)^2) + ((9.0*10^9)(+10^6 C)(+5 C)/(20000000 m)^2) + ((+5 C)(10000 m/s)(0.001 T)(1))/(10000 kg)]
a = 83537.5 / 10000
a = 8.35 N/kg

That is probably the final answer.