Let kinetic energy of a satellite is x

• • • • Derive the formula of the total energy of an orbiting body (or satellite) To calculate the total energy of an orbiting body (say a satellite) we have to find out the sum of its kinetic energy and potential energy. In one of our earlier posts, we have already seen The potential energy of an orbiting body can be expressed with the formula: PE = – GMm/r Here, the mass of the orbiting body is m and the radius of its orbit is r. M is the mass of a planet. Hence we can start with this now: Total Energy E = Potential energy + Kinetic energy = PE + KE = – GMm/r + (1/2)mv 2 ……….. (1) Now, as the body in orbit is in a circular motion, hence its centripetal force is being supplied by the gravitational force between mass m and M. mv 2/r = GMm/r 2 v 2 = GM/r ……………… (2) Hence, KE = (½) m v 2 = (½) m [GM/r] = (½) [GMm/r] …………………….(3) So, from (1) and (3): E = PE + KE = – GMm/r + (½) m v 2 = – GMm/r + (½) [GMm/r] = – (Gmm)/(2r) See also How to calculate the time the earth takes to go around the sun, using Newton's Universal Law of Gravitation? The total energy of an orbiting body or satellite E = – (Gmm)/(2r) ………………. (4) So equation 4 above is a formula giving the total energy of an orbiting body. It will always be negative because there is always more negative potential energy than positive kinetic energy for an orbiting body. Let’s do an example to see how this total energy changes in a change in orbit. Orbital Energy – graphs Figure 1 below shows three graphs representing the P...

Hint: We can recall that the period of revolution of an orbiting is the time taken to complete a cycle of its orbit. The centripetal force required to keep any object in orbit about a planet is provided by the gravitational force of attraction which pulls the object towards the planet Formula used: In this solution we will be using the following formulae; \[KE = \dfrac\]

The orbits of satellites about a central massive body can be described as either circular or elliptical. As mentioned earlier in work-energy relationship. Simply put, the theorem states that the initial amount of total mechanical energy (TME i) of a system plus the work done by external forces (W ext) on that system is equal to the final amount of total mechanical energy (TME f) of the system. The mechanical energy can be either in the form of potential energy (energy of position - usually vertical height) or kinetic energy (energy of motion). The work-energy theorem is expressed in equation form as KE i + PE i + W ext = KE f + PE f The W ext term in this equation is representative of the amount of work done by . For satellites, the only force is gravity. Since gravity is considered an , the W ext term is zero. The equation can then be simplified to the following form. KE i + PE i = KE f + PE f In such a situation as this, we often say that the total mechanical energy of the system is conserved. That is, the sum of kinetic and potential energies is unchanging. While energy can be transformed from kinetic energy into potential energy, the total amount remains the same - mechanical energy is conserved. As a satellite orbits earth, its total mechanical energy remains the same. Whether in circular or elliptical motion, there are no external forces capable of altering its total energy. Work and Energy Web Links Perhaps at this time you would like to use the links below to revie...

Identical satellites X and Y of mass m are in circular orbits around a planet of mass M. The radius of the planet is R. Satellite X has an orbital radius of 3R, and satellite Y has an orbital radius of 4R. The kinetic energy of satellite X is Kx. 1) In terms of Kx, the gravitational potential energy of the planet-satellite X system is (A) -2Kx (B) -Kx (C) -Kx/2 (D) Kx/2 (E) 2Kx 2) Satellite x is moved to the same orbit as satellite Y by a force doing work on the satellite. In terms of Kx, the work done on satellite X by the force is (A) -Kx/2 (B) -Kx/4 (C) 0 (D) +Kx/4 (E) +Kx/2 Show the work for each question. Answer: Option B and C are True. Explanation: The weight of the two blocks acts downwards. Let the weight of the two blocks be W. Solving for T₁ and T₂: w = T₁/cos 60° -----(1); w = T₂/cos 30° ----(2); equating (1) and (2) T₁/cos 60° = T₂/cos 30°; T₁ cos 30° = T₂ cos 60°; T₂/T₁ = cos 30°/cos 60°; T₂/T₁ =1.73. Therefore, option a is false since T₂ > T₁. Option B is true since T₁ cos 30° = T₂ cos 60°. Option C is true because the T₃ is due to the weight of the two blocks while T₄ is only due to one block. Option D is wrong because T₁ + T₂ > T₃ by simple summation of the two forces, except by vector addition. Answer:see below. Step-by-step explanation: To solve this problem, we can use the conservation of energy and conservation of momentum principles. Conservation of energy: The total initial energy is the rest energy of the proton and neutron, which is given by: Ei = ...

The following four statements about circular orbits are equivalent. Derive any one of them from first principles. • Negative kinetic energy equals half the potential energy ( − K=½ U). • Potential energy equals twice the total energy ( U=2 E). • Total energy equals negative kinetic energy ( E=− K). • Twice the kinetic energy plus the potential energy equals zero ( 2 K+ U=0). This is a key relationship for a larger problem in orbital mechanics known as the virial theorem. solution Circular orbits arise whenever the gravitational force on a satellite equals the centripetal force needed to move it with uniform circular motion. F c= F g mv 2 = Gm 1 m 2 r p r p 2 v 2= Gm 1 r Substitute this expression into the formula for kinetic energy. K=½ m 2 v 2 K=½ m 2 ⎛ ⎜ ⎝ Gm 1 ⎞ ⎟ ⎠ r K=½ Gm 1 m 2 r Note how similar this new formula is to the gravitational potential energy formula. K=+½ Gm 1 m 2 r U g=− Gm 1 m 2 r K=−½ U g The kinetic energy of a satellite in a circular orbit is half its gravitational energy and is positive instead of negative. When U and K are combined, their total is half the gravitational potential energy. E= K+ U g E=−½ U g+ U g E=½ U g E=− Gm 1 m 2 2 r The gravitational field of a planet or star is like a well. The kinetic energy of a satellite in orbit or a person on the surface sets the limit as to how high they can "climb" out of the well. A satellite in a circular orbit is halfway out (or halfway in, for you pessimists). Magnify practice problem 2 Start by dete...

I have a question about the concept of work in relation to conservative and non-conservative forces. Is Total work done $=$ change in kinetic energy or change in total energy $?$ This question arose when I was doing a question about the total work needed to move a satellite between orbits. To simplify things, I will use unrealistic values for $KE$ and $GPE$ but it will hopefully clarify my question. For example, a satellite has $GPE = -3$ $,$ $KE = 5$ in a lower orbit. It moves to a higher orbit with $GPE = -1$ $,$ $KE = 4$. My understanding is that the satellite does positive work converting fuel to $KE$ to instantaneously increase velocity (Hohmann transfer) i.e. raises $KE$ to $6$. Then as satellite moves off into higher orbit gravity does negative work converting its kinetic energy to potential energy, i.e. increases $GPE$ to $-1$. So in this scenario: Total change in energy $=$ $+1$ Change in $GPE $ $ =$ $+2$ Change in $KE$ $ =$ $-1$ In this scenario, what is the total work done? In the question I was doing, the answers stated that the total work done in moving the satellite to a higher orbit was the change in total energy. However, by the work energy theorem the work done should be equal to the change in kinetic energy which is $-1$. Also, does the presence of both conservative and non-conservative forces affect the calculation for total work done? In essence: what is the net work/total work done on the satellite, vs the work done by the satellite vs the work done by...