Earth has a radius of 6370km. North Korea claimed that they had successfully sen

Question

Earth has a radius of 6370km. North Korea claimed that they had successfully sent an ICMB at steep angle (i.e. near vertical direction) to a height of 2800km above ground. Assume the fuel of the rocket is completely in a short time, when the ICBM is still near the Earth's surface. (a) Can you still use U = mgy to calculate the potential energy of the ICBM in the case? Explain your answer. (b) Calculate the speed if the ICBM at the time when the fuel is burnt out. G = 6.674 times 10^-11 Nm^2kg^2 and the mass of earth is 5.972 times 10^24 kg. (c) What is the speed of the ICBM when it is at a height of 1400km?

Explanation / Answer

We can not use the relation U= mgy to calculate potential energy of the ICBM because the value of g does not remain constant for a range given in the problem. We use the formula for potential energy of ICBM as P E = GMm/R where M is mass of the earth, R distance from the center of the earth and G is universal gravitational constant.

total energy of moving ICBM at the surface = 1/2 m U2- GMm/R and total energy of the ICBM at the height of 2800 KM is 0 - GMm / (R+2800) Here kinetic energy at 2800 km is zero.so applying conservaation law of mechanical energy in conservative field we have 1/2 mU2 -GMm/R = - GMm/(R+2800) . Putting the value of G M and R from given problem we get the value of U as 6.18 km /sec Now we take the the case at the height of 1400 km. If V be the velocity at that height total energy of ICBM = 1/2 mV2-GMm/(R+1400) at this height.Equating it with the total energy at the surfaace, that is 1/2 mU2- GMm/R we can calculate the value of V, that is the value of velocity at the height

of 1400 km. It turns out to be 3.96 km/sec because the value of U is already known.

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