## Wednesday, September 14, 2011

### Solutions to Intermediate Astronomy Problems (6)- Part 2

We now conclude with the remainder of the problem set (4 -7) for Intermediate Astronomy Problems, Part (6). Once more, I give the problems then the solutions.

4) The orbital period of Jupiter's 5th satellite is 0.4982 days about the planet. Its orbital semi-major axis is 0.001207 AU. The orbital period and semi-major axis of Jupiter are 11.86 yrs. and 5.203 AU. Find the ratio of the mass of Jupiter to that of the Sun. Comment on why or why not you expect this result to be different from that in sample problem (1).

Solution:

The approach to solution is analogous to that we used in Tackling SIMPLE astronomy problems in terms of using a pair of ratios to apply to Kepler’s 3rd law, viz.

(P1/ P2)^2 = k(a1/ a2)^3

In this case we use the key equation at the end of Part (6), relating the period, T to u = G(m1 + m2).

T = 2π (a^3/u)^½

Now, in this problem we are dealing with Jupiter (planet) and its 5th satellite, with all the respective parameters given. Then we will have to pair two different ratios in a, T, i.e. of the satellite (moving around Jupiter) to Jupiter (around the Sun). We simply apply the same type of approach as shown with P1, P2 to a1, a2, thus:

For the planet and Sun:

T = 2π (a^3/u)^½

For the planet and satellite (Jupiter and its 5th satellite):

T' = 2π [(a')^3/u')^½

Then, dividing the the bottom form by the top form:

T'/T = {[(a')^3/ a ] [u/u']}^½

Or:

u/u' = (T'/T)^2 (a/a')^3

where u = (M + m) and u' = (m + m')

Given G (Newtonian grav. constant) cancels out, M= Sun's mass, m = Jupiter's mass, and m' = satellite's mass.

This is simplified by reducing a, T to AU and yrs, so that: a = 5.203 AU, a' = 0.001207 AU, T = 11.86 yrs. and T' = (0. 4982/365.25) yr = 0.00136 yr. Then:

(M + m)/(m + m') = (0.00136/11.86)^2 (5.203/0.001207)^3

Now the left side can be posed (by appropriate algebraic manipulation):

(M + m)/(m + m') = (M/m) [(1 + m/M)/(1 + m'/m)]

and in the limit of m/M <<< 1 we can write:

M/m = (0.00136/11.86)^2 (5.203/0.001207)^3 = 1.059 x 10^3

or: M = (1.059 x 10^3) m

So, the mass of the Sun is (1.059 x 10^3 Jupiter's) or inverting:

m/M = 1/(1.059 x 10^3) = 9.438 x 10^-4

The ratio of Jupiter's mass to that of the Sun.

We can check this accuracy by noting the mass of the Sun is given as 1.99 x 10^30 kg, so the mass of Jupiter would be:

m = (9.438 x 10^-4) (1.99 x 10^30 kg) = 1.878 x 10^27 kg

which compares to the (tabulated) mass of 1.89 x 10^27 kg (from info in sample problem (1). ) In effect, we obtain a much higher accuracy here and this likely is from using a ratio of two compounded masses (e.g. (M + m), (m + m') and the refinement provided by Newton's version of Kepler's 3rd law.

5) A communications satellite is in a circular equatorial orbit about Earth and always remains above a point of fixed longitude. If the sidereal day is 23h 56m long and the year 365.25 days in length and the distance of the satellite from Earth's center is 41,800 km, deduce the ratio of the mass of the Sun to the mass of the Earth. Hence, or otherwise, obtain the mass of Earth in kg if the mass of the Sun is as given in the check of the sample problem (1). (Take 1 AU = 149.5 x 10^6 km)

Solution:

The same exact approach obtains here as we saw for (4), the only difference is we're now considering the Earth-artificial satellite as a separate system in relation to the Earth and Sun - but their properties (a, T) will be similarly paired from Newton's version of the 3rd law of Kepler. Thus again:

u/u' = (T'/T)^2 (a/a')^3

where u = (M + m) and u' = (m + m')

Given G (Newtonian grav. constant) cancels out, M= Sun's mass, m = Earth's mass, and m' = artificial satellite's mass.

The satellite period T' = 23h 56m = 86 160s = (86160)/31 557 600 = 0.0027 yr.

and a' = (4.18 x 10^7 m)/ (1.495 x 10^11 m/AU) = 2.79 x 10^-4 AU

Then:

(M + m)/(m + m') = (0.0027/1)^2 (1/2.79 x 10^-4)^3

From (4) we found:

(M + m)/(m + m') = (M/m) [(1 + m/M)/(1 + m'/m)]

and in the limit of m/M <<< 1 we can write:

M/m = 3.4 x 10^5

The ratio of the mass of the Sun to the mass of the Earth.

Then the mass of the Earth will be:

m = (1.99 x 10^30 kg)/ (3.4 x 10^5) = 5.83 x 10^24 kg

(Compared to the tabulated value of 5.97 x 10^24 kg)

6) Based on the information you obtained from probs. 4 and 5 (as well as their solutions), compute the change in Jupiter's orbital period if it suddenly became the same mass as Earth.

Solution:

In this case, we relate the paired orbital parameters (a, T) of Jupiter to Earth, then make the appropriate mass change to solve for the new period. In this case, it helps to begin with the mass sums first.

We have:

u/u' = (T'/T)^2 (a/a')^3

where u = (M + m) and u' = (M + m'), the u applied for Jupiter and Sun, and the u' for Earth (m') and Sun (M). But we are demanding the new mass condition (Jupiter's mass = Earth's) or m = m', so:

u/u' = (M + m')/ (M + m') = 1

Therefore:

1 = (T'/T)^2 (a/a')^3

Let T' = 1 yr (Earth), a' = 1 AU (Earth), and we must then solve for the T, for Jupiter given its mass change to be equivalent to Earth's. We retain its semi-major axis, a = 5.203 AU. Then:

1 = (1/T)^2 (a/1)^3

Or:

T^2 = (a)^3 so T = [(a)^3)]^½ = [5.203^3]^½ = 11.868 yrs.

This compares to Jupiter's 11.856 yr. tabulated value to 5 significant figures. In other words, there is only a minor change in its period, of about 0.1%.

7) Halley's comet moves in an elliptical orbit of e= 0.9673. Calculate: a) the ratio of its linear velocities, and b) its angular velocities at perihelion and aphelion. Is it possible from this information to obtain the semi-major axis for this comet? If yes, then proceed to compute it!

Solution:

(a) The ratio is straightforward and the relation was given in the Blog:

(V_P/V_A) = (1 + e)/ (1 - e)

since we know e = 0.9673)

(V_P/V_A) = (1 + 0.9673)/ (1 - 0.9673) = 60.162

(b) This requires at least the semi-major axis be known, and no, there isn't enough to compute it from the information. But most people (at least amateur astronomers for whom these intermediate level problems are no big deal) know its period is 86 years. Then, using Kepler's third law:

a = [(86 yr)^2]^1/3 = 19.48 AU

This is then enough information to give V_P and V_A at least in terms of h (the specific relative angular momentum'). We need the actual mass of Halley's comet to obtain h itself.

So we have:

V_P = h/ a(1 - e) = h /[19.48 (1 - 0.9673)] = h/0.636 = 1.57 h

V_A = h/ a (1 + e) = h/ [19.48 (1 + 0.9673)] = h/38.3