A satellite is placed between the Earth and the Moon, along a straight line that connects their centers of mass. The satellite has an orbital period around the Earth that is the same as that of the Moon, 27.3 days. How far away from the Earth should this satellite be placed?

What you have here is a three-body problem. You cannot use Kepler's law to solve it.

It turns out that there is a point of unstable equilbrium between the earth and the moon where the gravity forces cancel and the satellite will remain at equilbrium at that location and revolve around the earth at the same angular velocity as the moon. There are also two other points along the earth-moon line (one beyond the moon and another on the opposite side of the earth from the moon) and two others at equilateral triangle points. This set of locations is called the Lagrangian points.

Put the satellite at a point between earth and moon where distances to earth and to the moon are such that
M(moon)/d(moon)^2 = M(earth)/d(earth)^2

But to solve this equation won't I need the distance of the moon from the satllite?

To determine the distance between the Earth and the satellite, we need to consider the gravitational forces acting on the satellite. The gravitational force between two objects is given by the equation:

F = G * (m1 * m2) / r^2

Where F is the gravitational force, G is the gravitational constant, m1 and m2 are the masses of the two objects, and r is the distance between the centers of mass of the two objects.

In this case, the gravitational force between the Earth and the satellite can be considered as centripetal force, as it is keeping the satellite in its orbit. The centripetal force is given by the equation:

F = (m * v^2) / r

Where m is the mass of the satellite, v is the orbital velocity, and r is the distance between the Earth and the satellite.

Since the satellite's orbital period around the Earth is the same as that of the Moon (27.3 days), we can use Kepler's Third Law, which states that the square of the orbital period is proportional to the cube of the semi-major axis of the orbit:

T^2 = (4 * pi^2 * r^3) / (G * M)

Where T is the orbital period, r is the semi-major axis, G is the gravitational constant, and M is the mass of the Earth.

In this case, the orbital period of the satellite is 27.3 days, so we substitute the values into the equation:

(27.3 days)^2 = (4 * pi^2 * r^3) / (G * M)

To make the calculations easier, we need to convert the orbital period from days to seconds:

(27.3 days)^2 = (4 * pi^2 * r^3) / (G * M)

(27.3 * 24 * 60 * 60)^2 = (4 * pi^2 * r^3) / (G * M)

Now we can solve for r by rearranging the equation:

r^3 = [(27.3 * 24 * 60 * 60)^2 * G * M] / (4 * pi^2)

Take the cube root of both sides:

r = [(27.3 * 24 * 60 * 60)^2 * G * M]^(1/3) / (4 * pi^2)^(1/3)

By plugging in the appropriate values for G (6.67430 × 10^-11 N m^2/kg^2), M (5.972 × 10^24 kg), and π (3.14159), we can calculate the distance r: