 # Gravitation: Potential

Physics

## Problems for Principle of Equivalence and Tides 2

Summary Problems for Principle of Equivalence and Tides 2

Problem : A rocket taking off from the earth is accelerating straight upwards at 6.6 m/sec2. How long will it take an apple of 0.2 kilograms to hit the floor of the rocket if it is dropped from a height of 1.5 meters?

The effective gravity in the spaceship is given by the gravity on earth plus the gravity due to the upward acceleration of the rocket: geff = 6.6 + 9.8 = 16.4 m/sec2. The time taken for an object to reach the ground can be determined from Galileo's kinematic equation which asserts that x = 1/2gt2, and thus t = =  0.43 secs. Of course the mass of the apple is irrelevant.

Problem : If you measure the speed of light on earth, will the result be the same as of you measured it in interstellar space, far from any gravitational fields?

Einstein's principle of equivalence demands that all measurements of the speed of light be the same. Imagine a spaceship in free-fall in a gravitational field, such that it is instantaneously at rest (it has not started to fall yet). There is effectively no gravity in these spaceships. The principle of equivalence demands that there be no method of determining whether they are falling or in a gravitational field, so it must be the case that an experiment to determine the speed of light will give the same result as if the experiment was performed far from any gravitational field.

Problem : If wood was found to fall at a different rate to plastic (ie.gwood gplastic ), what would be the consequences for the principle of equivalence?

If this was found to the true, the principle of equivalence would no longer hold. It was shown (see Inertial and Gravitational Masses that the gravitational mass was equivalent to the inertial mass if and only if gwood = gplastic, and that the same was true for all other materials.

Problem : A mass M is at the origin. Two masses m are at points (R, 0) and (R + x, 0) where x < < R. What is the difference in the gravitational force on the two masses? This is the longitudinal tidal force. (Hint: make some approximations)

The force is given by Newton's Universal Law: -   + =  -1 +  The second equality omitted the term in x2. Then using a binomial expansion we have: = (- 1 + (1–2x/R)) = Problem : Again, a mass M is at the origin. Now, two masses are at (R, 0) and (R, y), where y < < R. What is the difference in the gravitational force on the two masses, and what is its effect? This is the transverse tidal force.

To second order in (y/R), both masses are equally distant from the origin, and the magnitude of the force is essentially the same. The direction of the forces, however, differs in first order (y/R). In fact, this difference is the y-component of the force on the top mass: cosθ   = The difference points along the line joining the masses and acts to pull the masses together. The combination of longitudinal and tranverse tidal forces causes water on the side of the earth closest to the moon to be pulled towards it. Water on the opposite side from the moon is repelled (from the moon, causing it to bulge away from the earth, and water in between is pulled towards the center of the earth.