The percentage goals for AP Physics B and AP Physics C (Mechanics) are 6% and 18% respectively in the case of the topics under oscillations and gravitation. These topics were discussed earlier on this site. You can access those posts making use of the ‘search blog’ box.

Today we will discuss a few more multiple choice practice questions on gravitation. The following questions are useful for AP Physics B as well as AP Physics C aspirants:

(1) The velocity of escape from the earth’s surface is √(2GM

*/*R) and the work done by earth’s gravitational force to bring unit mass from infinity to a height*h*from the earth’s surface is GM/(R*+h*) where ‘G’ is the gravitational constant and ‘M’ is the mass of the earth. If a body of mass*m*is projected up with the escape velocity from the earth’s surface, how high will it rise? Neglect air resistance.(a) R/2

(b) R

(c) 2R

(d) 8R

(e) ∞

The velocity of escape is the velocity to be imparted to the body to make it escape from the gravitational pull (of the earth in the present case). So the height attained will be infinite [Option (e)].

(2) If the body in the above problem is projected up with 20 %

*of the escape velocity, how high will it rise?*(a) R/24

(b) R/20

(c) R/16

(d) R/8

(e) R/4

The kinetic energy supplied to the body is ½

*m*[0.2×√(2GM*/*R)]^{2}= 0.04 GM*m*/R. The above energy is used up in increasing the gravitational potential energy of the body. If the maximum height reached is

*h*, the increase in the gravitational potential energy is – GM*m/*(R+*h*) – (– GM*m/*R) = GM*m/*R – GM*m/*(R+*h*)[Note that the gravitational potential energy is negative]

Therefore, we have

0.04 GM

*m*/R = GM*m/*R – GM*m/*(R+*h*)Or, 0.04/R = 1/R – 1/(R+

*h*) =*h/*[R(R+*h*)]This gives

*h =*0.04(R+*h*) from which*h*= 0.04 R/0.96 =**R/24**.(3) The height of geostationary (synchronous) satellites above the surface of the earth is approximately 6

*R*where*R*is the radius of the earth. The orbital period of a research satellite at an altitude of 2.5*R*above the surface of the earth will be approximately(a) 16 hour

(b) 12√2 hour

(c) 8√2 hour

(d) 6√2 hour

(e) 6 hour

According to Kepler’s third law, the square of the orbital period of a satellite is directly proportional to the cube of the mean radius of the orbit of the satellite. Therefore, in the case of of geostationary (synchronous) satellites we have

24

^{2}α (7*R*)^{3}*since the orbital radius is 6**R*+*R =*7*R*and the orbital period is 24 hour.In the case of the research satellite the period

*T*is related to its orbital radius by*T*

^{2}α (3.5

*R*)

^{3}since its orbital radius is 2.5

*R*+

*R =*3.5

*R*

^{}

From the above relations (on dividing),

24/

*T*= 2^{3/2}Therefore,

*T =*24/2^{3/2 }= 24/(2√2) = 12/√2 = 6√2 hour.The following questions are meant solely for AP Physics C aspirants:

(4) Two satellites have orbital radii

*R*and 1.01*R*. Their orbital periods differ by(a) 1%

(b) 1.5%

(c) 2%

(d) 2.5%

(e) 3%

We have from Kepler’s third law,

*T*^{2}α*R*^{3}where*T*and*R*are the orbital period and orbital radius respectively.Therefore,

*T*=*kR*^{3/2}where*k*is the constant of proportionality.Taking logarithms, ln

*T =*ln*k +*(3/2) ln*R*Differentiating, d

*T/T*= 0 + (3/2) d*R/R*. This says that the fractional change in orbital period is 3/2 times the fractional change in orbital radius. In other words, the percentage change in orbital period is 3/2 times the percentage change in orbital radius. In the present problem the orbital radii differ by 1%. Therefore, the orbital periods will differ by (3/2)×1% = 1.5% [Option (b)].

(5) If the orbital radius of a satellite moving around the earth is increased by 2%, the orbital speed will

(a) remain unchanged

(b) decrease by 2%

(c) increase by 2%

(d) increase by 1%

(e) decrease by 1%

The orbital speed

*v*of a satellite moving around the earth in an orbit of radius*r*is given by*v =*√(

*GM/r*) where

*G*is the gravitational constant and

*M*is the mass of the earth.

[You can easily obtain this by equating the centripetal force to the gravitational pull:

*mv*^{2}/r =*GMm*/*r*^{2}where*m*is the mass of the satellite].This equation shows that the orbital speed will

*decrease*when the orbital radius is*increased*.Since

*G*and*M**are constants, we have* d

*v/v*= – ½ d*r/r*[You will get this by taking logarithms and by differentiating, as we did in question No.4].

This means that percentage increase in orbital speed = – (½)×percentage increase in orbital radius. The negative sign indicates the

*decrease*in the orbital speed because of the*increase*in orbital radius. Since the orbital radius of the satellite is

*increased*by 2%, the orbital speed is*decreased*by 1% [Option (e)].(6) A communication satellite of mass

*m*is placed initially in a temporary orbit of radius*r*_{1}around the earth_{.}How much work is to be done to shift it from this orbit to a permanent orbit of*greater*radius*r*_{2}?_{ }(Acceleration due to gravity at the sea level =*g*. Radius of the earth =*R*)(a)

*mgR*^{2}(r_{2}–*r*_{1})/*r*_{2}*r*_{1}(b)

*mgR*(r_{2}–*r*_{1})/ 2*r*_{2}(c)

*mgR*^{2}(r_{2}–*r*_{1})/ 2*r*_{2}*r*_{1}(d)

*mgR*^{2}(r_{2}–*r*_{1})/ 2*r*_{2}*r*_{1}(e) 2

*mgR*^{2}(r_{2}–*r*_{1})/*r*_{2}*r*_{1}Total energy (K.E. + P.E.) of a satellite of mass

*m*in an orbit of radius*r*is –*GMm/*2*r*where*M*is the mass of the earth.Work (

*W*) done in shifting the satellite from one orbit to the other is the difference between the energies in the orbits.Therefore,

*W =*–*GMm/*2*r*_{2}– (–*GMm/*2*r*_{1})[Note that the satellite has greater energy in the orbit of greater radius since the energy is negative by a smaller amount. When the radius is infinite, the energy is maximum and is equal to zero].

Thus

*W =*(*GMm/*2)(1/*r*_{1 }– 1/*r*_{2}) = (*GMm/*2)[(r_{2}–*r*_{1})/*r*_{2}*r*_{1}]Since

*g = GM/R*(as you can obtain from^{2}*mg = GMm/R*^{2}), we have*W = mgR*

^{2}(r

_{2}–

*r*

_{1})/ 2

*r*

_{2}

*r*

_{1}

## No comments:

## Post a Comment