Question

11)
Ch13. # 17
An artificial satellite circles the Earth in a circular orbit at a location where the acceleration due to gravity is 9.00 m/s2. Determine the orbital period of the satellite.

12)
Ch13. # 23
Comet Halley approaches the Sun to within 0.570 AU, and its orbital period is 75.6 yr. 1 AU = 1.50 x 1011 m). How far from the Sun will Halley

Answer 1

11)
g = GMe/Re^2, where Re = Radius of earth (6360km), G = 6.67x10^-11 Nm^2/kg^2, and Me = Mass of earth. On the earth's surface, g = 9.81 m/s^2, so the radius of your orbit is


R = Re * sqrt (9.81 m/s^2 / 9.00 m/s^2) = 6640km

here, the speed of the satellite is

v = sqrt(R*9.00m/s^2) = 7730 m/s

do the time it takes to make one pass is

T = 2*pi*R/v = 5397 s * 1h/3600s = 1.50 h

12)

Gravitational Paramater for the sun: mu_s = 1.32712 x10^11 km^3/s^2

Closest approach to sun = perihelion radius = r_p
r_p = 0.570 AU *1.496 x 10^8 km/AU
r_p = 8.5272x10^7 km

T = 75.6 yr *365.25 days/yr * 24 hrs/day * 60 min/hr * 60 s/min
T = 2.38575 x 10 ^ 9 seconds


From the equation for period you can find the semi-major axis a

T=2*pi*SQRT(a^3/mu_s)
==>
a= (mu_s*(T/(2pi))^2)^(1/3)
a= (1.32712x10^11 * (2.38575x10^9/(2*pi))^2)^(1/3)
a=2.67465 x 10^9 km

From the semi major axis a, you can find the aphelion radius r_a

a= (r_p + r_a)/2
==>
r_a = 2*a - r_p
r_a= 2 * (2.6465 x 10^9) - (8.5272 x 10^7)
r_a= 5.26403 x 10 ^9 km

In Astronomical Units
r_a= (5.26403 x 10 ^9) /(1.496 x 10^8)
r_a= 35.1874 AU

13)
Assuming the height is given in feet.

The pressure on the bottom will be

((h * 62.4) / 144) psig

The answer to b) will be
((h * 62.4) / 144) + ((mass of object) / ((area of tank in square feet) * 144)) psig

14)

? buoyant force
Balloon
?Weight

The buoyant force is equal to the weight of 325 m^3 of air.
Buoyant force = 325 * 1.20 * 9.8 = 3822 N

Total Weight = Weight of balloon and 325 m^3 of helium
Total Weight = 226 * 9.8 + 325 * 0.179 * 9.8 = 2784.915 N

Net force = buoyant force