w One Giant LeapThe escape speed from the surface of a slowly rotating air-less spherical planet is v. What is the minimum initial speed of a projectile launched from a pole that allows it to land on the equator? (SubmittedbyPhilipBlanco,Grossmont College,ElCajon,CA)

Solution: The situation is shown in Fig. 1. The pro-jectile is launched from the pole P and it lands at the equator E. The escape speed v from the surface of the planet is related with its mass M and its radius R; that relation is easily obtained from the total energy condi-tion E = 0 (parabolic orbit):

2 21 2 20 ,

2

GmM GM GME mv v v

R R R= − = ⇒ = ⇒ =

where m is the mass of the projectile and G is Newton’s gravitational constant.One possible orbit between P and E is a circular orbit, shown in Fig. 1; the projectile would travel parallel to the surface of the planet, with a speed v given by 2 2

22 2

0.71 .2

GmM v GM vm v

R R R

GM vv v v

R

′= ⇒ = = ⇒′

= = ≅ <′

The total energy in this case is negative:

21 10.

2 2 2

GmM GM GmM GmME mv m

R R R R= − = − = − <′

We want the projectile to go from P to E, launching it with the minimum initial speed. The corresponding orbit, of course, has to be elliptic since in that case the total energy will be also negative. The center C of the planet will be one of the two foci of the ellipse, as shown in Fig. 2.Since the angle PCE is 90o and CP

__ = CE

__ = R, by sym-

metry it follows that the radius vectors CP and CE

make an angle 45o = p/4 with the axis of the ellipse, as shown. We call V the initial speed of the projectile.

Physics Challenge for Teachers and Students

Boris Korsunsky, Column EditorWeston High School, Weston, MA 02493 korsunbo@post.harvard.edu

Solution to December 2012 Challenge

The Physics Teacher ◆ Vol. 50, 2012

The energy in an elliptic orbit is given by 2GmMaE =− ,

where a is the semi-major axis; so we can write 2

2 2

1

2 22

.

GmM GmMmV

R aGM GM GM

V vR a a

− = − ⇒

= − = −

This relation shows that the minimum V is reached when a is also minimum. Let us now use the equation of the ellipse in polar coordinates (r, j ), 2(1 ) ,

1 cosa ερ

ε ϕ−

=−

where e is the eccentricity. Applying it to point P, where r = R and j = p/4, we have 2

2

21(1 ) 2 ( ) ,121

2

aR a R f Rεε ε

εε

−−= ⇒ = ≡

−−

Fig. 1.

Fig. 2.

The Physics Teacher ◆ Vol. 51, 2013 173

Ramiro Moro (Cameron University, Lawton, OK)Jason L. Smith (Richland Community College, Decatur, IL)

Guidelines for contributors:

– We ask that all solutions, preferably in Word format, be submitted to the dedicated email addresschallenges@aapt.org. Each message will receive an auto-matic acknowledgment.

– The subject line of each message should be the same as the name of the solution file (see the instructions below).

– The deadline for submitting the solutions is the last day of the corresponding month.

– We can no longer guarantee that we’ll publish every suc-cessful solver’s name; each month, a representative selec-tion of names will be published, both in print and on the web.

– If your name is—for instance—Lady Gaga, please name the file “Mar13Germanotta” (do not include your first initial) when submitting the March solution.

– If you have a message for the Column Editor, you may contact him at korsunbo@post.harvard.edu; however, please do not send your solutions to this address.

As always, we look forward to your contributions and hope that they will include not only solutions but also your ownChallenges that you wish to submit for the col-umn.

Many thanks to all contributors and we hope to hear from many more of you in the future!

COLUMN EDITOR’S NOTE: There was an error in the published solution to the October “Challenge.” It was claimed that the wedge would acceler-ate at g/3 when the wedge angle approached 90°. However, the block would lose contact with the wedge at an angle of cos–1[(3– 5)/2] < 67.5°. This value can be obtained by set-ting the normal force between the block and wedge equal to zero in the system of equations in the published solution. Both Jason Smith and I wish to express our gratitude to Carl Mungan for pointing out the error.

so the minimum value of a is reached when the defined funtion f(e)is at a minimum. The condition f(e )= 0 gives function:

2

2 2

2 2 2

2

2 2(1 ) 1 ( 2 )2 2

0(1 )

2 2 2 22 2 0 2 02 2 2 2

2 2 8 42 2 1 0 2 1 2 1 0.41,2

ε ε ε

ε

ε ε ε ε ε

ε ε ε ε

− − − − − = ⇒

−

− + + − = ⇒ − + = ⇒

± −− + = ⇒ = = ± ⇒ = − ≅

because the eccentricity must satisfy |e| < 1. So

2

min 2

22 2 2min

min

2 2 2

min

2 21 ( 2 1) 1 1

2 2( 2 1)1 ( 2 1) 1 (2 1 2 2)

22 2 22 ( 2 1) 0.85 and

4 42( 2 1)

4 2

(2 2) 2 2

(2 2) ( 2 1)

so

2 1 0.64 .

a f R R R

R R R R

GM GM vv v v v

a R

v v v

v v v v v

− − − += − = =

− − − + −

+= = + = ≅−

= − = − = −+ +

= − − = −

= − ≅ < <′

The minimum initial speed of the projectile is

2 1 .v−

(ContributedbyFernandoFerreira,UniversidadedaBeiraInterior,Covilhã,Portugal)

We would also like to recognize the following successful contributors: Don Easton (Lacombe, Alberta, Canada)Norge Cruz Hernández (University of Seville, Seville,

Spain)Gerald. E. Hite (TAMUG, Galveston, TX)Art Hovey (Milford, CT)José Ignacio Íñiguez de la Torre (Universidad de Sala-

manca, Salamanca, Spain)Jarrett L. Lancaster (James Madison University,

Harrisonburg, VA)José Costa Leme (High School Lanheses, Viana do Cas-

telo, Portugal)Matthew W. Milligan (Farragut High School, Knoxville,

TN)