Question 1
Kinetic energy is postive and potential energy is negative.However the value of KE is half that of PE. In drawing the KE graph , draw on the positive axis but with half the value as seen in PE line. The change in speed cannot be determined directly from change in KE .Mathematically incorrect. Find the speed at the 2 different orbitals using the KE graph using calculation KE = 0.5mv2 and then find the change in speed.The answer is 520 ms-1 and not 1250 ms-1
Question 2
a) The force on satellite is gravitational force that point to the centre of the Earth , therefore the centre of circular motion is the centre of the Earth.
b) The orbit 200 km above the earths surface is not geostationary satellite . Hence the major mistake is to take period T=24 hours. Ans T: 5300s
c) the speed is v=rw or square root of GM/R where R = 6.6x106 m
d) The total energy is KE +PE and the formula should be -GMm/2R . Answer -6.1x109J
e) The negative sign implies that the satellite does not have enough energy to escape the gravitational field of Earth, and is therefore bound to the Earth.
f) Using total energy for this case will give a wrong answer because the satellite is at rest on the Earth surface. To put the satellite in orbit involves 2 different energy KE and PE. Sum up the increase in PE(3.8x108J) to increase in KE(6.1x109). The anwer is 6.48 x 10 9J
