Two Questions on Satellites

In the post (titled ‘Questions on Satellites’) dated August 30, 2006, two questions were discussed. The following question similar to the first question appeared in KEAM 2007 (Engineering) question paper:

A satellite is launched in a circular orbit of radius R around the earth. A second satellite is launched into an orbit of radius 1.01 R. The period of second satellite is longer than the first one (approximately) by

(a) 1.5% (b) 0.5% (c) 3%

(d) 1% (e) 2%

In the question discussed in the earlier post, the orbital radius was 1.02 R instead of 1.01R and the answer was obtained as 3%.

You may work out the present question to obtain the answer as 1.5%. If you have any difficulty, see the earlier post dated August 30, 2006. You may easily do this by clicking on the label satellite below this post.

Now see the following MCQ:

If the orbital radius of an artificial satellite is to be increased by 5%, the orbital speed will have to be

(a) decreased by 5% (b) increased by 5% (c) decreased by 2.5%

(d) increased by 2.5% (e) decreased by10%

The orbital speed is obtained by equating the centripetal force to the gravitainal pull:

mv²/r = GMm/r² where ‘m’ is the mass of the satellite, ‘v’ is the orbital speed, ‘r’ is the orbital radius, G is the gravitational constant and M is the mass of the earth.

From this v = √(GM/r).

Since G and M are constants, dv/v = – ½ dr/r

Instead of the fractional changes dv/v and dr/r, we can use percentage changes and write this equation as

percentage change in v = – ½ ×percentage change in r

Since the change in ‘r’ is an increment of 5%, the change in ‘v’ will be an increment of – ½ ×5%, which means a decrement of 2.5% [Option (c)].

All the essential points to be remembered in gravitation can be found at AP Physics Resources: Gravitation –Equations to be Remembered

Two Questions (MCQ) on Angular Momentum

The following two questions are similar in that both require the calculation of angular momentum in central field motion under inverse square law forces.

(1) An artificial satellite of mass ‘m’ is orbiting the earth of mass ‘M’ in a circular orbit of radius ‘r’. If ‘G’ is the gravitational constant, the orbital angular momentum of the satellite is

(a) [GMm²r]^1/2 (b) [GMmr]^1/2 (c) [GMm/r]^1/2 (d) [GMm²/r²]^1/2 (e) [GMm²r³]^1/2

The orbital angular momentum of a satellite is mvr where ‘v’ is the orbital speed. [Angular momentum = Iω = mr²ω = mr²v/r = mvr where ‘I’ is the moment of inertia and ‘ω’ is the angular velocity of the satellite].

The centripetal force required for the circular motion of the satellite is supplied by the gravitational pull so that we have

mv²/r = GMm/r²

From this, m²v²r² = GMm²r so that angular momentum mvr = [GMm²r]^1/2

(2) In a hydrogen atom in its ground state, the electron of mass ‘m’ is moving round the proton in a circular orbit of radius ‘r’. The orbital angular momentum of the electron is (with usual meaning for symbols)

(a) [m²e²r/4πε₀]^1/2 (b) [m²er/4πε₀]^1/2 (c) [me²r²/4πε₀]^1/2

(d) [me²r/4πε₀]^1/2 (e) [me²r³/4πε₀]^1/2

The steps for finding the orbital angular momentum of the electron are similar to those in question No.1, with the difference that the centripetal force is supplied in this case by the electrostatic attractive force between the proton and the electron.

We have mv²/r = (1/4πε₀)e²/r² from which m²v²r² = (1/4πε₀) ×me²r, so that orbital angular momentum, mvr = [me²r/4πε₀]^1/2

Questions on Satellites

The following questions on satellites are simple, but slightly different from the common direct questions. See whether you can solve them yourself. Go through the solution given, only after your trial.
(1)A satellite is moving in an orbit of radius ‘r’ around the earth. A second satellite is moving in an orbit of radius (1.02)r. Then the orbital period of the second satellite is greater than that of the first by approximately
(a) 0.2% (b) 2% (c) 3% (d) 10.2% (e) 6%
We have T² α r³ (Kepler’s law) from which T α r^3/2. Since the increment in the orbital radius ‘r’ is 0.02r, the fractional increment in the orbital period ‘T’ is (∆T)/T = (3/2)(∆r)/r = (3/2) × 0.02r/r = 0.03. The percentage increase is 0.03×100 = 3% [option (c)].
(2) If the orbital period of a satellite is ‘T’, its kinetic energy is directly proportional to
(a) T (b) T^-1 (c) T^3/2 (d) T^2/3 (e) T^-2/3

We have, kinetic energy, E = GMm/2r where G is the gravitational constant, M is the mass of the earth ‘m’ is the mass of the satellite and ‘r’ is the radius of the orbit. Therefore, E α 1/r. Since T² α r³ in accordance with Kepler’s law, we can write r α T^2/3. Therefore, E α T^-2/3. The correct option is (e).