# Convert constant of gravitation to days and AUs [duplicate]

I'm working on a problem with celestial bodies and for my purpose days and AUs are more appropriate units than seconds and meters. So I tried to convert the constant of gravitation, $G$, like this:

$$k= \frac{\frac{\text{AU}^3}{\text{kg}\times \text{D}^2}}{\frac{\text{m}^3}{\text{kg}\times \text{s}^2}}=(\frac{\text{AU}}{\text{m}})^3(\frac{s}{D})^2=\frac{(1.496\times 10^{-11})^3}{(24\times 60\times 60)^2}=4.485\times 10^{-43}$$ $$G=6.673\times10^{-11}\times k = 6.673\times10^{-11}\times4.485\times 10^{-43}=2.993\times 10^{-53}$$

Now I want to do some calculations with this value. First of all, what is the orbital period of a circular orbit around the sun?

$$v_0=\sqrt{\frac{G M_\odot}{r}}$$ $$t = \frac{2\pi r}{v_0}$$

The initial velocity has been adapted so that the centripetal force equals the force of gravitation, so the orbit will be circular.

For $r=1$ I expected that the orbital period would be a couple of hundred days, since it takes about 365 days for the earth to circle the sun and one AU is the distance between the sun and the earth. However, when I do the calculation I get that $t=8.14462\times 10^{11}$. What did I do wrong?

• For the benefit of future readers: The Google Calculator is very good at unit conversion. Also, it's better to use the standard gravitational parameter $\mu = GM$, when possible, since it's more accurate than separate values of $G$ & $M$. Commented Jul 10, 2018 at 8:25

You have a couple of mistakes. First, if you say that $G = 6\times 10^{-11} k$, then $k$ should be $\frac{\text{m}^3}{\text{kg}\cdot \text{s}^2}$. If we instead define $k$ as you did, then it is a dimensionless number: the conversion factor between the two sets of units. You should rather have said that $6\times 10^{-11} / k$ is the value of $G$ in the system of units you want to use. Note that we're dividing by $k$ instead of multiplying; this is because $\frac{\text{m}^3}{\text{kg}\cdot \text{s}^2} = \frac1{k} \frac{\text{AU}^3}{\text{kg}\cdot \text{D}^2}$.

Your other problem is that $\text{AU}/\text{m}$ should have $10^{11}$ instead of $10^{-11}$. Taking that into account, we get that $k = 4.485\times 10^{23}$, and $G = 1.488\times 10^{-34} \frac{\text{AU}^3}{\text{kg}\cdot \text{D}^2}$. WolframAlpha then says that $t = 365.13 \text{D}$, which seems pretty close to correct.

G = 6.673 × 10 −11 N*m^2/kg^2

(Freedman, Roger; Geller, Robert M.; Kaufmann, William J.. Universe (Page A6). W. H. Freeman. Kindle Edition.)

N = kg*m/s^2, so the units of G as defined above is (kg * m^3)/(kg^2 * S^2) or m^3/(kg * S^2)

To convert from seconds to days: 60s/min * 60min/hr * 24hr/day = 86,400 sec/day

and from meters to AU: 1 AU = 1.4960 × 10^11 m

(Freedman, Roger; Geller, Robert M.; Kaufmann, William J.. Universe (Page A6). W. H. Freeman. Kindle Edition.)

G = 6.673 × 10 −11 (kg * m^3)/(kg^2 * S^2) = 6.673 × 10 −11 (kg * m^3)/(kg^2 * S^2)* ((1 AU / 1.4960 × 10^11 m)^3) / ((1 d / 86400 s)^2) = 6.673 × 10 −11 * (6.68445 x 10^-12)^3 / (1.1574 x 10^-05)^2 = 6.673 × 10 −11 * 2.9868x10^-34 / 1.33959x10^-10 = 6.673 × 10 −11 * 2.2296 x 10^-24 G = 1.4878 x 10^-34 AU^3 /(kg*d^2)

Then you make an assumption of circular motion.

To do so the following equations apply (Giancoli Douglas. General Physics (Page 88) Prentice-Hall, 1984):

a = v^2 / r , acceleration of a particle moving with tangential velocity, v, in circle of radius, r.

g = G * (m / r^2), acceleration of a gravity, with mass of body causing acceleration, m, at distance r

a = g and v^2 / r = G * (m / r^2)
thus [1] v^2 = G * (m / r), which your first equation.

Then you use a form of the average velocity equation: (Giancoli Douglas. General Physics (Page 15, 90) Prentice-Hall, 1984)

v = x1 - x0 / t1 - t0 = x1 / t1 for x0 = 0, t0 = 0 x1 = 2*(3.14)r for the distance around the circumference of a circle t1 = T, the revolution period thus v = 2(3.14)r / T or [2] T = 2(3.14)*r / v

Solving [1]

G = 1.4878 x 10^-34 AU^3 /(kg*d^2) M = 1.989 × 10^30 kg (Freedman, Roger; Geller, Robert M.; Kaufmann, William J.. Universe (Page A6). W. H. Freeman. Kindle Edition) r = 1 AU

v = SQRT[(1.4878 x 10^-34 AU^3 /(kg*d^2)) * 1.989 × 10^30 kg / 1 AU] = SQRT[2.9593 x 10^-04] = 1.72 x 10^-02 AU/day

Then using the velocity to solve [2]

T = 2 * (3.14) * 1 AU / 1.72 x 10^-02 AU/day T = 3.65246 X 10^02 days

However, you can go further if you substitute [2] in [1]: [1] v^2 = G * (m / r) [2] v = 2*(pi)*r / T

(4 * (pi)^2 * r^2) / T^2 = G * (m / r) [3] (4*pi^2)/(G*m) = T^2 / r^3

Thus any body orbiting a mass, m, will have its period and orbital distance related by a constant, (4*pi^2)/(G*m).

This is core of Kepler's third law, which allows you to relate the orbital distance and period of any two bodies orbiting the same mass with the following equation:

[4] T1^2 / T2^2 = r1^3 / r2^3

• Hi Kirk Carver, welcome to physics.SE! Please note that this site supports mathjax. Click here for a tutorial. We encourage people to use it here. Thanks! Commented Jul 6, 2018 at 23:55