Final answer:
To find the distance from the center of Eqrgp where the Lunatik's speed and direction remain constant, we would need to equalize the gravitational forces from Eqrgp and its moon, assuming gravitational equilibrium. However, without the Lunatik's mass and additional details, we cannot solve for this distance.
Step-by-step explanation:
To determine how far from the center of Eqrgp the Lunatik is at the time when its speed and direction are constant, we would need to consider the gravitational forces acting upon it from both Eqrgp and its moon.
As per the provided information, the Lunatik is in a position where it experiences the gravitational pull from both bodies equally but opposite, causing no change in velocity. This state is known as gravitational equilibrium.
To solve this, we use the formula for gravitational force, F = G * (m1 * m2) / r^2, where G is the gravitational constant, m1 and m2 are the masses of the two objects, and r is the distance between them.
We set the forces due to gravity from both Eqrgp and the moon to be equal, which gives us M1/r1^2 = M2/r2^2, where M1 and M2 are the masses of Eqrgp and its moon, respectively, and r1 and r2 are the distances from the respective bodies to the Lunatik. We know that r1 + r2 = 3.31 x 10⁹ m (the distance between the centers of Eqrgp and its moon).
However, without the mass of the Lunatik and further details regarding the forces or the specific point of equilibrium, we cannot calculate the exact distance. Such calculations typically involve solving for r1 (or r2) and then using the total distance to find the Lunatik's position.
Generally, this sort of problem would use the concept of the center of mass or the Lagrange points where an object can maintain its position relative to two larger bodies.