We're now at the point of tackling true elliptical orbits, as opposed to elliptical approximations. So all the problems set or done will reflect that. We will use the diagram shown, which denotes a true elliptical orbit defined by some radius vector, r. This is the variable distance from central object, i.e. Sun, to the orbiting body, i.e. planet, at any given time. It is different from the semi-major axis, a, because the latter is a defined mean while the former depends on the actual distance at some time t. Thus, what this instalment is all about is generalizing the treatment of orbits to be able to work more and varied problems, including those involving planetary satellites and artificial Earth satellites.
Now, let the points A and P denote the aphelion (farthest point) and perihelion (closest point) respectively and V_A, V_P the respective velocities at those orbit extrema. As may be deduced here, points A and P are the only ones in the whole orbit for which the velocities are truly tangential or at right angles to the radius vectors for those positions. Consequently, we can write:
V = (2π/T) r
where r is the radius vector at the point, and T is the period.
If Kepler's 2nd law holds at every point (equal areas swept out in equal intervals of time) we also have:
r^2 (2π/T) = h
where 'h' is a constant ('specific relative angular momentum') which is twice the rate of area description (i.e. by the radius vector). Thus, if the radius vector is r1, then h = 2A1, when A1 = π(r1)^2. Hence, at aphelion and perihelion only we have:
V = h/r
For the perihelion velocity we have:
V_P = h/ a(1 - e)
where a is the semi-major axis, and e is the eccentricity.
For the velocity at aphelion:
V_A = h/ a (1 + e)
Then the ratio of velocities is:
(V_P/V_A) = (1 + e)/ (1 - e)
The correct energy equation can be written:
½V^2 - u/r = C
where C is an energy integration constant.
Since for any bound system of masses m1 and m2, u = G(m1 + m2), where G is the Newtonian gravitational constant (G = 6.7 x 10^-11 Nm^2/kg^2) then if we know V_P and V_A, along with a and e, we can compute C, viz.
C = ½V_P^2 - u/a(1 - e)
at perihelion, and
C = ½V_A^2 - u/a(1 + e)
at aphelion.
subtracting the lower equation from the upper and using the ratio of velocities expression (V_P/V_A), we get:
(V_P)^2 = u/a [{1 + e)/ (1 - e)]
and:
(V_A)^2 = u/a [{1 - e)/ (1 + e)]
Combining these, we obtain, much more simply:
(V_A)(V_P) = u/a
with further manipulation using the energy equation, the ambitious reader should be able to obtain:
V^2 = u (2/r - 1/a)
which is also known as the "vis viva" equation, one of the most important in celestial mechanics.
Now, what if the planet orbit is fully circular, so r = a?
Then we see the vis viva reduces to:
V^2 = u (2/a - 1/a) = u [(2a - a)/a^2] = u (a/a^2) = u/a
But as we saw:
V = 2πa/T
then:
T = 2π (a^3/u)^½
which holds even when the orbit is elliptical.
Sample Problems:
1) You've discovered a new Jovian moon that takes 30 days to go around Jupiter and is 19 Jupiter diameters from the planet. Deduce Jupiter's mass based on this information?
Solution *(One way - there are others!)
This problem basically uses Kepler's 3rd or 'harmonic law' to obtain Jupiter's mass from Jupiter's moon's motion - knowing the mass of the Sun (1 solar mass). From the harmonic law we know that the period squared is proportional to the semi-major axis of the orbit cubed:
T^2 ~ a^3
Where T is in years and a in astronomical units or AU. (1 AU = 1.496 x 10^8 km) Note: when we use AU and yrs. we are implicitly comparing the elements of the unknown system (e.g. moon going round Jupiter) with the known system of the Earth going round the Sun. For Earth, we know T = 1 year, a = 1 AU.
For the newly discovered moon going round Jupiter, T(m) = 30 days. Or, in the units we need:P(m) = 30 days/ 365 days/year = 0.0822 yrs.
The distance or semi-major axis for the moon's orbit is 19 Jupiter diameters, or a(m) = 19 x (1.43 x 10^5 km) = 2.717 x 10^6 km where the quantity in brackets is the known equatorial diameter of Jupiter, found from a table of planetary data.
The value above is a(m) = 0.018 AU since 2.717 x 10^6 km)/ (1.496 x 10^8 km/ AU) = 0.018 AU. Now, according to the use of the 3rd law, the relations between period and semi-major axis obtaining for the Earth-Sun ought to be in the same proportion as the relation between period and semi-major axis for the moon of Jupiter and Jupiter. Thus, we can write the proportion:
M(S)/ M(J) = 1/ (a^3/ T^2)
Now, M(S) is the mass of the Sun, and M(J) is the mass for Jupiter. The numerator on the right side is simply the ratio of a^3 to T^2 for Earth :
(1 AU)^3/ (1 yr.) ^2 = 1 AU^3/yr^2
The denominator is for the moon of Jupiter's values, or:
(a^3)/ T^2 = (0.018 AU)^3/ (0.0822 yr)^2= 0.00086 AU^3/yr^2
Now, put all these values back into the proportion we obtained:
M(S)/ M(J) = 1/ 0.00086
Note that the units of AU and yrs. cancel out between the numerator (e.g. values for Earth or 1 AU^3/yr^2) and the denominator (values for moon of Jupiter, e.g. (0.018 AU)^3/ (0.0822 yr)^2)
What we have then at the end, is a simple proportion from which we can find the mass of Jupiter (M(J). Cross multiplying and setting the terms equal:
M(J) = 0.00086 M(S)
In other words, the mass of Jupiter is about 8.6 x 10^-4 of the Sun's.
Or, since the Sun = 1 solar mass (by definition) then Jupiter must be 0.00086 solar masses.
To check this, consult a table of planetary masses to find Jupiter's mass = 1.89 x 10^27 kg and the Sun's = 1.99 x 10^30 kg. Then taking the ratio:
(1.89 x 10^27 kg)/ (1.99 x 10^30 kg) = 0.0094, or slightly off.
(2) The Pluto-Charon system, i.e. the orbit of Charon with respect to Pluto, has an eccentricity e = 0.0020. The semi-major axis of the orbit is 19, 450 km. The mass of Pluto = 1.27 x 10^22 kg and the mass ratio (Charon to Pluto) is found to be m(c)/m(P) = 0.12. From this information, find:
a) The mass of Charon
b) The ratio of the velocity of Charon at perihelion to aphelion
c) The period of Charon, and its velocity
Solution:
a) The mass is found using the mass ratio. Since m(c)/m(P) = 0.12 and m(P) =1.27 x 10^22 kg , then:
m(C) = 0.12 (1.27 x 10^22 kg ) = 1.52 x 10^21 kg
b) This ratio is obtained from:
(V_P/V_A) = (1 + e)/ (1 - e)
and we know, e = 0.0020, so:
(V_P/V_A) = (1 + 0.0020)/ (1 - 0.0020) = (1.0020)/ (0.998)
(V_P/V_A) = 1.004
c) The period is obtained from: T = 2π (a^3/u)^½
where a = 19, 450 km = 1.94 x 10^7 m
For preserving unit consistency this is one time we decline to use units of AU and years, and instead use the distance in meters to conform with the required units for G. Then:
T = 2π [(1.94 x 10^7 m)^3/ [6.7 x 10^-11 Nm^2/kg^2) (1.422 x 10^22 kg)]^½
T = 5.5 x 10^5 s = 6.37 days
The velocity is:
V = 2πa/T = 2π( 1.94 x 10^7 m )/(5.5 x 10^5 s)= 221.6 m/s
Other Problems:
1) Derive the vis viva equation using three or more of the equations given in this blog. (Hint: all incorporate u, and two incorporate e)
2) The Earth's aphelion distance is 1.01671 AU and its perihelion distance is 0.98329 AU. Given its eccentricity e = 0.016, then use this information and any other (from previous blogs) to find:
a) the velocities at aphelion and perihelion
b) the energy constants C(A) and C(P) at each of these points, and h, the 'specific relative angular momentum'.
c) Hence or otherwise, use the vis viva equation to confirm the results you obtained in (a)
3) Attempt to obtain an improved value for Jupiter's mass (from what the first sample problem yields) using: T = 2π (a^3/u)^½ (Recall u = G(m1 + m2) and Jupiter's mass and the Sun's have already been given along with G, in the blog general information)
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).
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)
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.
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!
(Answers will be given in two sets)
Now, let the points A and P denote the aphelion (farthest point) and perihelion (closest point) respectively and V_A, V_P the respective velocities at those orbit extrema. As may be deduced here, points A and P are the only ones in the whole orbit for which the velocities are truly tangential or at right angles to the radius vectors for those positions. Consequently, we can write:
V = (2π/T) r
where r is the radius vector at the point, and T is the period.
If Kepler's 2nd law holds at every point (equal areas swept out in equal intervals of time) we also have:
r^2 (2π/T) = h
where 'h' is a constant ('specific relative angular momentum') which is twice the rate of area description (i.e. by the radius vector). Thus, if the radius vector is r1, then h = 2A1, when A1 = π(r1)^2. Hence, at aphelion and perihelion only we have:
V = h/r
For the perihelion velocity we have:
V_P = h/ a(1 - e)
where a is the semi-major axis, and e is the eccentricity.
For the velocity at aphelion:
V_A = h/ a (1 + e)
Then the ratio of velocities is:
(V_P/V_A) = (1 + e)/ (1 - e)
The correct energy equation can be written:
½V^2 - u/r = C
where C is an energy integration constant.
Since for any bound system of masses m1 and m2, u = G(m1 + m2), where G is the Newtonian gravitational constant (G = 6.7 x 10^-11 Nm^2/kg^2) then if we know V_P and V_A, along with a and e, we can compute C, viz.
C = ½V_P^2 - u/a(1 - e)
at perihelion, and
C = ½V_A^2 - u/a(1 + e)
at aphelion.
subtracting the lower equation from the upper and using the ratio of velocities expression (V_P/V_A), we get:
(V_P)^2 = u/a [{1 + e)/ (1 - e)]
and:
(V_A)^2 = u/a [{1 - e)/ (1 + e)]
Combining these, we obtain, much more simply:
(V_A)(V_P) = u/a
with further manipulation using the energy equation, the ambitious reader should be able to obtain:
V^2 = u (2/r - 1/a)
which is also known as the "vis viva" equation, one of the most important in celestial mechanics.
Now, what if the planet orbit is fully circular, so r = a?
Then we see the vis viva reduces to:
V^2 = u (2/a - 1/a) = u [(2a - a)/a^2] = u (a/a^2) = u/a
But as we saw:
V = 2πa/T
then:
T = 2π (a^3/u)^½
which holds even when the orbit is elliptical.
Sample Problems:
1) You've discovered a new Jovian moon that takes 30 days to go around Jupiter and is 19 Jupiter diameters from the planet. Deduce Jupiter's mass based on this information?
Solution *(One way - there are others!)
This problem basically uses Kepler's 3rd or 'harmonic law' to obtain Jupiter's mass from Jupiter's moon's motion - knowing the mass of the Sun (1 solar mass). From the harmonic law we know that the period squared is proportional to the semi-major axis of the orbit cubed:
T^2 ~ a^3
Where T is in years and a in astronomical units or AU. (1 AU = 1.496 x 10^8 km) Note: when we use AU and yrs. we are implicitly comparing the elements of the unknown system (e.g. moon going round Jupiter) with the known system of the Earth going round the Sun. For Earth, we know T = 1 year, a = 1 AU.
For the newly discovered moon going round Jupiter, T(m) = 30 days. Or, in the units we need:P(m) = 30 days/ 365 days/year = 0.0822 yrs.
The distance or semi-major axis for the moon's orbit is 19 Jupiter diameters, or a(m) = 19 x (1.43 x 10^5 km) = 2.717 x 10^6 km where the quantity in brackets is the known equatorial diameter of Jupiter, found from a table of planetary data.
The value above is a(m) = 0.018 AU since 2.717 x 10^6 km)/ (1.496 x 10^8 km/ AU) = 0.018 AU. Now, according to the use of the 3rd law, the relations between period and semi-major axis obtaining for the Earth-Sun ought to be in the same proportion as the relation between period and semi-major axis for the moon of Jupiter and Jupiter. Thus, we can write the proportion:
M(S)/ M(J) = 1/ (a^3/ T^2)
Now, M(S) is the mass of the Sun, and M(J) is the mass for Jupiter. The numerator on the right side is simply the ratio of a^3 to T^2 for Earth :
(1 AU)^3/ (1 yr.) ^2 = 1 AU^3/yr^2
The denominator is for the moon of Jupiter's values, or:
(a^3)/ T^2 = (0.018 AU)^3/ (0.0822 yr)^2= 0.00086 AU^3/yr^2
Now, put all these values back into the proportion we obtained:
M(S)/ M(J) = 1/ 0.00086
Note that the units of AU and yrs. cancel out between the numerator (e.g. values for Earth or 1 AU^3/yr^2) and the denominator (values for moon of Jupiter, e.g. (0.018 AU)^3/ (0.0822 yr)^2)
What we have then at the end, is a simple proportion from which we can find the mass of Jupiter (M(J). Cross multiplying and setting the terms equal:
M(J) = 0.00086 M(S)
In other words, the mass of Jupiter is about 8.6 x 10^-4 of the Sun's.
Or, since the Sun = 1 solar mass (by definition) then Jupiter must be 0.00086 solar masses.
To check this, consult a table of planetary masses to find Jupiter's mass = 1.89 x 10^27 kg and the Sun's = 1.99 x 10^30 kg. Then taking the ratio:
(1.89 x 10^27 kg)/ (1.99 x 10^30 kg) = 0.0094, or slightly off.
(2) The Pluto-Charon system, i.e. the orbit of Charon with respect to Pluto, has an eccentricity e = 0.0020. The semi-major axis of the orbit is 19, 450 km. The mass of Pluto = 1.27 x 10^22 kg and the mass ratio (Charon to Pluto) is found to be m(c)/m(P) = 0.12. From this information, find:
a) The mass of Charon
b) The ratio of the velocity of Charon at perihelion to aphelion
c) The period of Charon, and its velocity
Solution:
a) The mass is found using the mass ratio. Since m(c)/m(P) = 0.12 and m(P) =1.27 x 10^22 kg , then:
m(C) = 0.12 (1.27 x 10^22 kg ) = 1.52 x 10^21 kg
b) This ratio is obtained from:
(V_P/V_A) = (1 + e)/ (1 - e)
and we know, e = 0.0020, so:
(V_P/V_A) = (1 + 0.0020)/ (1 - 0.0020) = (1.0020)/ (0.998)
(V_P/V_A) = 1.004
c) The period is obtained from: T = 2π (a^3/u)^½
where a = 19, 450 km = 1.94 x 10^7 m
For preserving unit consistency this is one time we decline to use units of AU and years, and instead use the distance in meters to conform with the required units for G. Then:
T = 2π [(1.94 x 10^7 m)^3/ [6.7 x 10^-11 Nm^2/kg^2) (1.422 x 10^22 kg)]^½
T = 5.5 x 10^5 s = 6.37 days
The velocity is:
V = 2πa/T = 2π( 1.94 x 10^7 m )/(5.5 x 10^5 s)= 221.6 m/s
Other Problems:
1) Derive the vis viva equation using three or more of the equations given in this blog. (Hint: all incorporate u, and two incorporate e)
2) The Earth's aphelion distance is 1.01671 AU and its perihelion distance is 0.98329 AU. Given its eccentricity e = 0.016, then use this information and any other (from previous blogs) to find:
a) the velocities at aphelion and perihelion
b) the energy constants C(A) and C(P) at each of these points, and h, the 'specific relative angular momentum'.
c) Hence or otherwise, use the vis viva equation to confirm the results you obtained in (a)
3) Attempt to obtain an improved value for Jupiter's mass (from what the first sample problem yields) using: T = 2π (a^3/u)^½ (Recall u = G(m1 + m2) and Jupiter's mass and the Sun's have already been given along with G, in the blog general information)
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).
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)
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.
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!
(Answers will be given in two sets)
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