Calculating Speed and Time of Nolan Ryan's Baseball Orbit on the Moon

[JJ89]

Suppose that Nolan Ryan stands on the surface of the moon and throws a baseball horizontally. If the baseball has a high enough speed and does not strike any mountain, it can orbit around the moon and, after completing the orbit, strike Nolan from behind. The mass of the moon is mm = 7.35×(10*22power kg), and its radius is rm = 1740 km. The gravitation constant G = 6.67×10(-11power) Nm2/kg2.

a) Find the speed at which Nolan must throw the ball for such a circular orbit? V= ms

(b) How long (in hours) does the ball take to complete one orbit? T= hrs


is the equation V=G*R? if so i used 7.35*10to the 22nd power * 6.67 *10 to -11 power?
i got 4.90245e12, and it said i was wrong.

The Attempt at a Solution

 

[JJ89]

but i can't figure out the equation for B

 

[JJ89]

did i do A right?

 

[kuruman]

is the equation V=G*R? if so i used 7.35*10to the 22nd power * 6.67 *10 to -11 power?
i got 4.90245e12, and it said i was wrong.

No, the equation is not V = G*R. Look up the correct equation for satellite motion in a circular orbit in your textbook. You can also derive it using Newton's Second Law combined with Newton's Law of gravitation.

 

[rwisz]

:

I would approach this problem by using the equation for centripetal force, which is F = (mv^2)/r, where m is the mass of the object, v is its velocity, and r is the radius of the circular orbit. In this case, the centripetal force is provided by the gravitational force between the baseball and the moon, which can be calculated using the formula F = (G*m*m)/r^2, where G is the gravitational constant, m is the mass of the moon, and r is the distance between the baseball and the center of the moon.

To find the speed at which the baseball must be thrown for a circular orbit, we can equate these two equations and solve for v:

(G*m*m)/r^2 = (mv^2)/r

Solving for v, we get v = √(G*m/r), where G = 6.67×10^-11 Nm^2/kg^2, m = 7.35×10^22 kg, and r = 1740 km = 1.74×10^6 m.

Plugging in these values, we get v = √((6.67×10^-11 Nm^2/kg^2)*(7.35×10^22 kg)/(1.74×10^6 m)) = 1639.6 m/s.

Therefore, the speed at which Nolan must throw the ball for it to orbit around the moon is approximately 1639.6 m/s.

To find the time it takes for the ball to complete one orbit, we can use the formula for the period of a circular orbit, which is T = 2πr/v, where r is the radius of the orbit and v is the orbital velocity. In this case, the radius of the orbit is the radius of the moon, which is 1.74×10^6 m, and the velocity is the orbital velocity we just calculated, 1639.6 m/s.

Plugging these values into the equation, we get T = 2π*(1.74×10^6 m)/(1639.6 m/s) = 6.74×10^5 s.

To convert this to hours, we divide by 3600 seconds per hour, giving us T = 6.74×10^5 s/3600 s/hr = 187.2 hours.