Question

Task 2 (Gravitational Potential of the Earth) In good approximation, the earth can be regarded as a sphere of homogenous mass density with radius R and total mass M. The gravitational potential of a mass m which has a distance r to the center of the earth is given by GMm , r>R U(r) 2R where Eo and G are constants a) Calculate the force F(r) which acts on a mass m at the earth's surface Hint: The gradient of a radial symmetric function f(r) is given by b) In reality the earth is additionally rotating with a constant angular velocity w around a axis through both poles. Consider a person standing on the equator. Which pseudo-forces are acting on the person? c) How long would one day need to be such that the person on the equator (see figure) has no weight? d) Neglect now the earth's rotation. What is the minimal velocity necessary in order for a vertically upwards starting non-accelerating probe to leave the gravitational field and explore space? Achieved Points:

Answer 1

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a) U(r) = {= GMm/r: r > R GMm/2R r^2/R^2 + E_0, r < R concept: Force F(r) = -nabla^vector U(r) = -dU(r)/dr r^vector calculation: F(r) = -du/dr = d/dr(GMm/r) [for r > R F(r) = -GMm/r^2 e_r^^ F(R) = -GMm/R^2 e_r^^ Here -ve sign implies the force is attractive. (b) Due to rotation of earth two pseudo-forces will act on the person. (i) Centrifugal force. (ii) coriolis Force \r\n (c) At the equator effective gravity is given by g^' = g - R omega^2 where R rightarrow radius of earth. For the person to be masses g^' = 0 R omega^2 = g omega = squareroot g/R. since T = 2pi/omega T = 2pi squareroot R/g = 2pi squareroot 6.4 times 10^6/10 = 2pi squareroot 64 times 10^4 = 2pi times 8 times 10^2 sec = 5026 sec T = 1.396 Hr Hence days will be almost 1.4 Hr, long. (d) let Minimum velocity be V. then in order