Calculating the Mass of Saturn Using Orbital Data

  • Thread starter Thread starter Fysicsx
  • Start date Start date
  • Tags Tags
    Mass Saturn
AI Thread Summary
The discussion centers on calculating the mass of Saturn using Titan's orbital data, specifically its 16-day orbital period and a radius of 1,222,000 km. The user initially misapplied the formula for orbital period, incorrectly dividing the radius by two. The correct approach involves using the full orbital radius in the calculations. The gravitational constant is confirmed as 6.67e-11 m^3*kg^-1*s^-2. The user acknowledges the mistake and expresses gratitude for the assistance.
Fysicsx
Messages
3
Reaction score
0
titan, a moon of saturn, has a 16 day orbital period and an orbital radius of 1,222,000 km. what is the mass of saturn?

my attempt:

t = 2*pi*r^(3/2)/sqrt(g*m)
t = 16 days = 1,382,400 s
r = 1,222,000/2 km = 611,000,000 m

1,382,400 = 2*pi*611,000,000^(3/2)/sqrt(6.67e-11*m) => m = 7.065e25 kg

the answer should be 5.6e26 kg, what am i doing wrong?
 
Physics news on Phys.org
Fysicsx said:
titan, a moon of saturn, has a 16 day orbital period and an orbital radius of 1,222,000 km. what is the mass of saturn?

my attempt:

t = 2*pi*r^(3/2)/sqrt(g*m)
t = 16 days = 1,382,400 s
r = 1,222,000/2 km = 611,000,000 m

1,382,400 = 2*pi*611,000,000^(3/2)/sqrt(6.67e-11*m) => m = 7.065e25 kg

the answer should be 5.6e26 kg, what am i doing wrong?
Hello Fysicsx. Welcome to PF !

Where does the following equation come from?

t = 2*pi*r(3/2)/√(g*m)

What is g in this equation?
 
hi sammys,

thanks for the warm welcome

f_g = g*m_1*m_2/r^2 (law of gravitation)
v = sqrt(g*m/r) (circular orbit)

v = 2*pi*r/t
t = 2*pi*r/v = 2*pi*r*sqrt(r/(g*m)) = 2*pi*r^(3/2)/sqrt(g*m) (circular orbit)

g is the gravitational constant 6.67e-11 m^3*kg^-1*s^-2
 
Fysicsx said:
hi sammys,

thanks for the warm welcome

f_g = g*m_1*m_2/r^2 (law of gravitation)
v = sqrt(g*m/r) (circular orbit)

v = 2*pi*r/t
t = 2*pi*r/v = 2*pi*r*sqrt(r/(g*m)) = 2*pi*r^(3/2)/sqrt(g*m) (circular orbit)

g is the gravitational constant 6.67e-11 m^3*kg^-1*s^-2
You divided r by 2.

Solve this for r, not v : v = sqrt(g*m/r)
 
oops that was careless lol. thanks for your help, sammys.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top