Imagine a parallel universe in which the electric force has the same properties as in our universe but there is no gravity. In this parallel universe, the sun carries charge Q, the earth carries charge −Q, and the electric attraction between them keeps the earth in orbit. The earth in the parallel universe has the same mass, the same orbital radius, and the same orbital period as in our universe.
Calculate the value of Q. (Consult Appendix F as needed.)
Express your answer using two significant figures.
Not having appendix F I'll ty to talk you through it. First you need the gravitational force between Earth and the sun. Get this from F = GmM/r^2. Next just use Coulomb's Law
F (the one you just found) = kQ^2/r^2.
In fact you can save a step and ignore the distance between them with
Q = sqrt(GMm/k).
G = 6.67e-11, k = 9e9, m = 5.98e24 but I don't know the mas of the sun offhand
To calculate the value of Q in the given scenario, we can use the equation for the electric force between two charged objects:
Fe = k * |Q1 * Q2| / r^2
Where:
- Fe is the electric force between the objects
- k is the electrostatic constant (in our universe, it is approximately 9 x 10^9 N m^2/C^2)
- Q1 and Q2 are the charges of the objects
- r is the distance between the objects
In the given scenario, the electric attraction between the sun and the earth replaces the gravitational force. Since the earth is in orbit, the electric force is equal to the centripetal force needed to keep the earth in its orbit. The centripetal force is given by:
Fc = (m * v^2) / r
Where:
- Fc is the centripetal force
- m is the mass of the earth
- v is the orbital velocity of the earth
- r is the orbital radius
In both universes (our universe and the hypothetical parallel universe), the mass, orbital radius, and orbital period are the same for the earth. Therefore, the only difference is the type of force involved—gravity in our universe and electric force in the parallel universe.
To equate the electric force and the centripetal force in the parallel universe, we can set them equal to each other:
Fe = Fc
Then plug in the respective equations:
k * |Q1 * Q2| / r^2 = (m * v^2) / r
Since the earth carries charge -Q and the sun carries charge Q:
k * |-Q * Q| / r^2 = (m * v^2) / r
Now we can simplify the equation:
k * Q^2 / r^2 = m * v^2 / r
Solving for Q:
Q^2 = (m * v^2 * r) / (k * r^2)
Q^2 = (m * v^2) / (k * r)
Q = √[(m * v^2) / (k * r)]
To find the value of Q, we need to know the values of m (mass of the earth), v (orbital velocity of the earth), r (orbital radius), and k (the electrostatic constant). Since these values are not provided in the given question, we are unable to perform the calculation to find the value of Q.
To solve for the value of Q in the parallel universe, we need to equate the electric force between the sun and the earth to the centripetal force keeping the earth in its orbit.
In our universe, the centripetal force is given by:
F_centr = m * (v^2 / r)
where m is the mass of the earth, v is its orbital velocity, and r is the orbital radius.
In the parallel universe, the electric force between the sun and the earth is given by:
F_elec = k * (Q^2 / r^2)
where k is the electrostatic constant and Q is the charge of the sun in the parallel universe.
Equating these two forces, we get:
F_centr = F_elec
m * (v^2 / r) = k * (Q^2 / r^2)
Rearranging the equation, we have:
Q^2 = (m * v^2 * r) / (k * r^2)
Now, let's consider the given values:
m = mass of the earth (same as in our universe)
v = orbital velocity of the earth (same as in our universe)
r = orbital radius of the earth (same as in our universe)
k = electrostatic constant (same as in our universe)
We can plug in these values and calculate Q:
Q^2 = (m * v^2 * r) / (k * r^2)
Since the mass, orbital velocity, and orbital radius remain the same as in our universe, the value of Q in the parallel universe will also be the same as it is in our universe.
Therefore, the value of Q in the parallel universe will be the same as the charge of the sun in our universe.