- #1

- 12

- 0

- Thread starter TheSerpent
- Start date

- #1

- 12

- 0

- #2

gneill

Mentor

- 20,879

- 2,839

Relevant Equations? Attempt at Solution?

- #3

- 12

- 0

Unsure on where to begin!

- #4

gneill

Mentor

- 20,879

- 2,839

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

- #5

- 12

- 0

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.

- #6

gneill

Mentor

- 20,879

- 2,839

Okay, that's progress.

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.

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 [itex] \vec{a} = \vec{F}/m [/itex]

- #7

- 12

- 0

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.

- #8

gneill

Mentor

- 20,879

- 2,839

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.

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.

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.

- #9

- 12

- 0

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.

- Replies
- 12

- Views
- 3K

- Last Post

- Replies
- 5

- Views
- 610

- Replies
- 5

- Views
- 1K

- Replies
- 4

- Views
- 497

- Replies
- 10

- Views
- 697

- Last Post

- Replies
- 7

- Views
- 3K

- Last Post

- Replies
- 1

- Views
- 5K

- Last Post

- Replies
- 1

- Views
- 1K

- Last Post

- Replies
- 8

- Views
- 3K

- Last Post

- Replies
- 1

- Views
- 2K