- #1

- 1

- 0

The Apollo spacecraft must be traveling at what velocity in order to remain in a 110 kilometer orbit around the moon? The magnitude of the Apollo’s orbital velocity can be computed from the laws of gravitational motion. When the Apollo spacecraft is orbiting the moon it has an orbital velocity that is directed along a tangent to the circular orbit, Vt. A velocity in this direction would take the Apollo out of its orbit around the moon if it was not counteracted by the gravitational pull of the moon on the spacecraft. The moon’s gravitational force acts on the Apollo spacecraft pulling it in toward the center of the moon which is also the center of the orbital path. A force that is directed along a radius to a circular path and points toward the center of the circle is sometimes referred to as a “Centripetal Force”. Because the gravitational force of the moon is keeping the Apollo in circular motion and is always directed toward the center of the circular orbit it is the “Centripetal Force” in this question.

We can use the following Physics Equations to solve this problem: Equation 1: Newton’s Law of Gravitation Fg = (G x m1 x m2) / (r2) WHERE: Fg, force of gravity of moon in Newtons, N G, universal gravitational const.= 6.6732 x 10-11 N∙m2/kg2 m1 is the mass of the moon in kg m2 is the mass of the Apollo spacecraft in kg r, separation distance from center of m1 to m2 center in m

Equation 2: Centripetal acceleration of any object in uniform circular motion can be proven through geometry, but is not proven here.

actr = vt 2/r

WHERE: actr = Radially and center directed acceleration, centripetal acceleration, caused by Fg, in m/s2 vt = Tangentially directed velocity of the Apollo spacecraft in m/s r = Radius of the circular path, distance from center of the circle to center of the Apollo

Equation 3:Newton’s Second Law Fctr = m2 actr WHERE: Fctr = Radially and center directed force, centripetal force, in Newtons m2 = Mass of object traveling in circle, Apollo spacecraft, in kg actr = Radially and center directed acceleration, centripetal acceleration, m/s2

Equation 4: Substituting equation 2 into equation 3, we get, Fctr = m2∙vt 2/r Because in this question the gravitational force IS the radially and center directed force, Fctr,

Equation 1 and 4 can be set equal to each other: Equation 1=Equation 4: Fg = Fctr (G ∙m1∙m2)/r2= m2∙vt 2/r Solving this for the linear tangential velocity of the Apollo spacecraft yields:

Equation 5:Orbital velocity of any object in uniform circular motion around the moon vt = (G ∙m1/r)1/2 WHERE: vt = Tangentially directed linear velocity of the Apollo spacecraft in m/s G = The universal gravitational constant = 6.6732 x 10-11 N∙m2/kg2 m1 = The mass of the moon in kg r = Radius of the circular path in meters from the center of the moon to the center of the Apollo spacecraft

Use Equation 5 to calculate the answer to question 1.

Find the volumetric mean radius of the moon and then add that to the orbital altitude of the Apollo to calculate R. Don't forget that R in the equation needs to be in meters!

Question 2: Calculate the escape velocity of an object from the moon. The escape velocity (vesc) of a body is the initial velocity required to go from an initial point in a gravitational field to a point infinitely far away and with a final velocity of zero. The Law of Conservation of Energy can be used to calculate the escape velocity under these conditions. What velocity would an object need in order to escape from the gravitational field of the moon?

EQUATION 1: Conservation of Mechanical Energy Σ E initial=Σ E final Sum of all initial energy=Sum of all final energy E kinetic, initial + E grav,initial=E kinetic, final + E grav,final E kinetic, initial + E grav,initial=0+0

EQUATION 2: Kinetic Energy Formula E kinetic=½ m ∙vesc2 WHERE: E kinetic = kinetic energy of escaping object in Joules, J m = mass of the escaping object, in kg Vesc = escape velocity from the moon in m/s

EQUATION 3: Gravitational Energy E gravitational= - G∙M∙m /r WHERE: E grav = potential gravitational energy of the moon G, universal grav. const.= 6.6732 X 10-11 N∙m2/kg2 M = the mass of the object in kg m = the mass of the Moon in kg r = Distance from Moon’s center to object on moon's surface, or mean radius of Moon in m

EQUATION 4: Escape Velocity Substituting Equation 2 and 3 into Equation 1 yields an equation you can use to solve for the escape velocity of the moon: E kinetic, initial + E grav,initial=0 E kinetic, initial= - E grav,initial ½ m ∙vesc2=G∙M∙m /r vesc=(2 G∙m /r)1/2 WHERE: vescis the escape velocity from the moon in m/s G, universal gravit. const= 6.6732 x 10-11 N∙m2/kg2 m = the mass of the moon in kg r = Mean tadius of the moon in m

Notice that the escape velocity of an object does NOT depend on the mass or size of the object trying to escape! Use equation 4 to calculate the escape velocity for an object on the moon.