A spacecraft of mass is first brought into an orbit around the earth. The earth (together with the spacecraft) orbits the sun in a near circular orbit with radius ( is the mean distance between the earth and the sun; it is about 150 million km).
(a) What is the speed (in m/s) of the earth in its orbit of radius around the sun with a mass
We want the spacecraft to fall into the sun. One way to do this is to fire the rocket in a direction opposite to the earth's orbital motion to reduce the spacecraft's speed to zero (relative to the sun).
(b) What is the total impulse that would have to be given by the rocket to the spacecraft to accomplish this? You may ignore the effect of the earth's gravitation as well as the orbital speed of the spacecraft around the earth as the latter is much smaller than the speed of the earth around the sun. Thus, you may assume that the spacecraft, before the rocket is fired, has the same speed in its orbit around the sun as the earth. Express your answer in terms of and (enter m for and v_0 for ).
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We will now show that there is a more economical way of doing this (i.e., a much smaller rocket can do the job). By means of a brief rocket burn the spacecraft is first put into an elliptical orbit around the sun; the boost is provided tangentially to the earth's circular orbit around the sun (see figure). The aphelion of the new orbit is at a distance from the sun. At aphelion the spacecraft is given a backward impulse to reduce its speed to zero (relative to the sun) so that it will subsequently fall into the sun.
(c) Calculate the impulse required at the first rocket burn (the boost). Express your answer in terms of , and (enter I_0 for , R for and r for ).
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(d) What is the speed of the spacecraft at aphelion? Express your answer in terms of , and (enter v_0 for , R for and r for ).
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(e) Calculate the impulse required at the second rocket burn (at aphelion). Express your answer in terms of , and (enter I_0 for , R for and r for ).
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(f) Compare the impulse under b) with the sum of the impulses under c) and e) (i.e find ), and convince yourself that the latter procedure is more economical. Express your answer in terms of , and (enter I_0 for , R for and r for ).
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(a) To find the speed of the earth in its orbit around the sun, we can use the formula for the orbital velocity:
v = sqrt(GM/r)
where v is the velocity, G is the gravitational constant (approximately 6.67x10^-11 m^3/kg/s^2), M is the mass of the sun (approximately 1.989x10^30 kg), and r is the radius of the orbit.
Given that r is the mean distance between the earth and the sun (150 million km or 150x10^9 m), we can calculate the speed:
v = sqrt((6.67x10^-11 m^3/kg/s^2 * 1.989x10^30 kg) / (150x10^9 m))
Calculating this expression will give us the speed of the earth in its orbit around the sun.
(b) To calculate the total impulse required to reduce the spacecraft's speed to zero relative to the sun, we need to find the change in momentum. Since impulse is defined as the change in momentum, we can use the formula:
Impulse = ∆p = m∆v
where ∆p is the change in momentum, m is the mass of the spacecraft, and ∆v is the change in velocity.
In this case, the change in velocity is the speed of the earth in its orbit around the sun, since we want the spacecraft to match that velocity. So:
Impulse = m * v
Substituting the given values, we can calculate the total impulse required.
(c) To calculate the impulse required for the first rocket burn (the boost), we can use the principle of conservation of angular momentum. Since the rocket provides a tangential thrust, the angular momentum remains constant.
Angular momentum = mvr = m'v'r'
where m is the mass of the spacecraft, v is the velocity before the boost, r is the distance from the sun to the spacecraft before the boost, m' is the mass of the spacecraft after the boost, v' is the velocity after the boost, and r' is the distance from the sun to the spacecraft at aphelion.
Given that the aphelion distance is r + R, where R is the distance of the aphelion from the circular orbit, we can calculate the impulse required at the first rocket burn by rearranging the equation:
Impulse = m'v' - mvr
Substituting the given values, we can calculate the impulse.
(d) To find the speed of the spacecraft at aphelion, we can use the conservation of mechanical energy. At aphelion, the gravitational potential energy is zero and the kinetic energy is equal to the negative of the potential energy at the circular orbit.
0.5m'v'^2 - GMm' / (r + R) = -GMm' / r
Simplifying this equation and solving for v', we can find the speed of the spacecraft at aphelion.
(e) To calculate the impulse required at the second rocket burn (at aphelion), we can use the same principle of conservation of angular momentum as in part (c).
Impulse = m''v''
where m'' is the mass of the spacecraft at aphelion and v'' is the final velocity after the second rocket burn.
Substituting the given values, we can calculate the impulse.
(f) To compare the impulse under part (b) with the sum of the impulses under parts (c) and (e):
Total Impulse = Impulse from part (b) - (Impulse from part (c) + Impulse from part (e))
Substituting the given values, we can calculate the total impulse.
Note: The calculations provided above are only a general guide on how to approach the problem. Specific numerical values will vary based on the actual numbers provided in the question.
To calculate the speed of the earth in its orbit around the sun (a), we need to use the formula for the centripetal acceleration:
a = v^2 / r
where a is the centripetal acceleration, v is the orbital speed, and r is the radius of the orbit.
To find the speed of the earth, we need to find the radius of its orbit. The mean distance between the earth and the sun is given as 150 million km, which is 150 x 10^6 km or 150 x 10^9 m. This is the radius (r) of the earth's orbit around the sun.
Substituting the values into the equation, we have:
v^2 = a * r
The centripetal acceleration is caused by the gravitational force between the sun and the earth, and is given by the equation:
a = GM / r^2
where G is the gravitational constant (6.674 x 10^-11 Nm^2/kg^2) and M is the mass of the sun.
The mass of the sun (M) can be approximated as 1.989 x 10^30 kg.
Substituting the values into the equation for acceleration, we have:
a = (6.674 x 10^-11 Nm^2/kg^2 * 1.989 x 10^30 kg) / (150 x 10^9 m)^2
Now we can substitute the value of acceleration into the equation for speed:
v^2 = a * r
v^2 = [(6.674 x 10^-11 Nm^2/kg^2 * 1.989 x 10^30 kg) / (150 x 10^9 m)^2] * (150 x 10^9 m)
Simplifying and taking the square root gives us the speed (v) of the earth in its orbit around the sun.