Suppose it takes 3 days for a space vehicle to travel from the Earth to the moon. About how long would it take the same vehicle traveling a the same speed to reach Neptune?
Simplistically, at 240,000/3 = 80,000 mpd, it would take 2,789,089,173/80,000 = 34,864 days = 95.45 years, on average, 2,789,089,173 miles being the average distance between Earth and Neptune.
More realistically:
How long does it take to get to the planet Neptune? (I've looked all over the internet and in encyclopedias). Thanks
Let me give you some information regarding the times to travel to all the planets, including Neptune, then a bit more detail regarding the trip time to Pluto. I haven't written up an in depth version for Neptune but i think you will learn from the Pluto trip information.
The time for a space probe, launched from Earth, to reach any planet, is primarily a function of the distance, final velocity, and velocity direction at burnout of the rocket stage. There are two extremes to examining the time required to travel to a planet in a direct planet to planet flight. One requires a minimal expenditure of rocket energy but results in the longest trip time. The other requires a huge expenditure of rocket energy but results in a much shorter trip time, relatively speaking.
The minimum energy approach for a probe to reach any planet on its own is by means of the Hohmann Transfer Orbit (HTO). By minimum energy, I mean the lowest possible final velocity of the probe as it departs its earth orbit. The Hohmann Transfer Orbit is an elliptical orbit that is tangent to both of the orbits of the planets between which the transfer is to be made. In other words, a probe placed into a heliocentric orbit about the Sun would leave the influence of the earth with a velocity vector tangent to the earth's orbit and arrive at the destination planet's orbit 180 degrees away from the departure point, and with a velocity vector tangent to its orbit.
The following table spells out the times it takes to reach each planet via a 180 degree transit, elliptical Hohmann Transfer Orbit, using the mean distances between the earth and the planets. The times to return would be exactly the same using the same type of trajectory.
Planet.............M........V..........E...........M.........J.........S...........U..........N........P
Time----------.289-----.40---------------.709....2.732...6.05…...16.12....30.6....45.5...years
.....................105.....146....................259..............................................................days
Mean dist---.387----.723---------..----1.524---5.20----9.54-----19.18---30.06--39.5.....A.U.*
*A.U.=astronomical unit=mean distance between sun and earth=~92,960,240 miles.
Another way of looking at a grand tour trip is to launch inwards toward Mercury on a HTO trajectory. Upon arrival, fire the onboard rocket to propel the vehicle on another HTO from Mercury to Venus. Upon arrival at Venus, fire up again, and propel the vehicle on yet another HTO, from Venus to Mars, and so on until the vehicle reaches Pluto.
The individual trip times for this approach would be as follows:
Earth to Mercury--105 days
Mercury to Venus--75.5 days
Venus to Mars--217.5 days
Mars to Jupiter--3.08 years
Jupiter to Saturn--10 years
Saturn to Uranus--27.2 years
Uranus to Neptune--61 years
Neptune to Pluto--102.5 years--Total trip time--205 years. Too long and all the planets would have to be in a very specific orientaion to start out with, the likelihood of happening being totally remote.
We can get to any of the planets quicker by a higher velocity, more direct route, requiring more rocket energy, or by the use of more selective planetary swingbys when planetary orientations are in our favor.
Pluto Trip
<< How long will it take to travel from earth to Pluto> >>
About 45.5 years on an elliptical, minimum energy, Hohmann Transfer trajectory to the mean distance but you could get there in as little as 2.5 years on a direct high energy hyperbolic trajectory. Of course, anything in between is possible also depending on the size of the launch rocket used. Let me try to explain.
The time for a space probe, launched from Earth, to reach the planet Pluto, or any planet for that matter, is primarily a function of the distance, final velocity, and velocity direction at burnout of the rocket stage. There are two extremes to examining the time required to travel to a planet in a direct planet to planet flight. One requires a minimal expenditure of rocket energy but results in the longest trip time. The other requires a huge expenditure of rocket energy but results in a much shorter trip time, relatively speaking.
The minimum energy approach for a probe to reach any planet on its own is by means of the Hohmann Transfer Orbit. By minimum energy, I mean the lowest possible final velocity of the probe as it departs its earth orbit. The Hohmann Transfer Orbit (HTO) is an elliptical orbit that is tangent to both of the orbits of the planets between which the transfer is to be made. In other words, a probe placed into a heliocentric orbit about the Sun would leave the influence of the earth with a velocity vector tangent to the earth's orbit and arrive at the destination planet's orbit 180 degrees away from the departure point, and with a velocity vector tangent to its
orbit.
Lets explore what it takes to send a space probe to our planet in question. Lets assume a launch vehicle has already placed our probe and its auxiliary rocket stage in a circular, 250 mile high, low Earth orbit (LEO) with the required orbital velocity of 25,155 feet per second, fps., (17,147 MPH).
By definition, a probe being launched on a journey to another planet, must be given sufficient velocity to escape the gravitational pull of Earth. A probe that is given exactly escape velocity, will depart the Earth on a parabolic trajectory, and just barely escape the gravitational field. This means that its velocity will be approaching zero as its distance from the center of the Earth approaches infinity. If however, we give our probe more than minimal escape velocity, it will end up on a hyperbolic trajectory and with some finite residual velocity as it approaches infinity. This residual velocity that the probe retains is called the "hyperbolic excess velocity." When added to the velocity of the Earth in its orbit about the Sun, the result is the heliocentric velocity required to place the probe on the correct Hohmann transfer trajectory to rendezvous with Pluto.
It is worth noting at this point that it is somewhat naive to talk about a space probe reaching infinity and escaping a gravitational field completely. It is somewhat realistic, however, to say that once a probe has reached a great distance (on the order of 500,000 to one million miles) from Earth, it has, for all intensive purposes,
escaped. At these distances, it has already slowed down to very near its hyperbolic excess velocity. It has therefore become convenient to define an imaginary sphere surrounding every gravitational body as the body's "sphere of influence", SOI. When a space probe passes through this SOI, it is said to have truly escaped. Over the years, it has become difficult to get any two people to agree on exactly where the SOI should be located but, nevertheless, the ficticious boundry is widely used in preliminary lunar and interplanetary trajectory studies.
First, an interesting aspect of going to Pluto is its highly elliptical orbit around the sun. At its closest approach to the sun, the perihelion, Pluto is ~2,762,731,000 miles from the sun. At its most distant point, the aphelion, it is ~4,607,283,000 miles from the sun. Quite a difference. As stated earlier, a probe that departs from a 250 mile high LEO with the minimal escape velocity of 36,605 fps (24,270 MPH) will end up at the edge of the Earth's SOI with near zero velocity and remain there forever. We must then give our probe a final Earth departure velocity in excess of minimal escape velocity. Lets back into the required velocity to reach Pluto at its mean distance from the sun, 3,685,000,000 miles, in the following manner.
The final heliocentric velocity (on a solar orbit) required by our probe for a Hohmann transfer to Pluto's perihelion is ~135,900 fps (92,638 MPH), relative to the Sun. Since the probe, at the beginning of its journey, picks up and retains the velocity of the Earth about the Sun, 97,700 fps (66,600 MPH), we can subtract this from our required heliocentric velocity to determine our required hyperbolic excess velocity. Thus, 136,900 - 97,700 = 38,200 fps, the hyperbolic excess velocity required by our probe at the edge of the Earth's SOI. To achieve this hyperbolic excess velocity, on the hyperbolic escape trajectory from our 250 mile parking orbit, requires a final rocket burnout velocity of 52,220 fps, relative to the Earth. The velocity vector must nominally be perpendicular to the radius from the Sun, through the center of the Earth and parallel to the Earth's direction around the Sun. Since our circular parking orbit velocity is already 25,155 fps, we need therefore only add a velocity increment (deltaV) of 27,065 fps with our auxiliary rocket stage burn.
Our launch vehicle has already delivered us to our 250 mile LEO. The auxiliary stage now fires, imparting its impulsive velocity change of 27,065 fps to push us out of the circular orbit onto our required hyperbolic orbit from which it escapes the Earth altogether. Unfortunately, the journey would take ~45.5 years to travel the 180 degrees around the Sun and arrive in the vicinity of Pluto with a velocity of ~4,573 fps (3,118 MPH). (It would take ~30 years to reach its perihelion and ~63 years to reach its aphelion) Obviously, the phasing of the planets is important for this type of transfer and the launch must occur at a carefully predetermined time such that both Pluto, and the probe, will reach the heliocentric rendezvous vicinity at the same time.
Clearly one can reach the planet quicker by launching the probe on a hyperbolic heliocentric trajectory but it would require the expenditure of considerably more energy to do so, meaning bigger rockets and higher costs. Just for example, lets arbitrarily double the deltaV imparted to our probe from the 250 mile parking orbit. This would result in a final burnout velocity of 2(27,065) + 25,155 = 79,285 fps. which in turn results in a hyperbolic excess velocity of 70,840 fps. Our final heliocentric departure velocity from Earth's orbit around the Sun is then 70,840 + 97,700 = 168,540 fps. This places our probe on a fast, direct, hyperbolic heliocentric trajectory toward Pluto with a rendezous time of ~2.5 years from departure. (Of course, we would now launch our payload at a time when Pluto was directly in front of the earth, heliocentrically speaking, since we are no longer bound by the Hohmann Transfer constraints of having to rendevous with Pluto 180 degrees after departing earth.) But at what price? In fact, by doubling the impulsive deltaV of our auxilliary stage, we can no longer accomplish it with a single stage rocket. We would, in fact, have to now use a three stage rocket out of earth orbit, weighing more than 200,000 pounds, to reach the required velocity. Quite a price for getting there faster. And don't lose sight of the fact that the launch vehicle that lifted us up into LEO in the first place, also has to grow by a considerable amount. Clearly, there are other, less expensive, choices of departure velocities for shortening trip times. The choice is primarily a function of how much money one is willing to spend on a launch vehicle and the launch site hardware.
Now all of the above assumes the probe is totally on its own for the entire journey. Another approach to reaching another planet is to launch the probe toward an intermediary planet, say Jupiter, or even Venus, for instance, and let the planet alter the trajectory of the probe through what is typically referred to as the sling shot, swingby, or gravity turn approach.The use of the term slingshot is somewhat misleading however as it tends to imply that the intermediate planet throws, or hurls, the probe away from it toward another planet with added energy or velocity relative to the turning planet, which it does not. In reality, a probe that enters the sphere of influence of another planets gravitational field, will be pulled in toward the planet along a hyperbolic trajectory to some predetermined minimum altitude, called the perigee altitude, increasing its velocity all the way, and once passing this minimum altitude point, will continue out and away from the planet on the exact mirror image trajectory that it came in on. When the probe again reaches the edge of the planets sphere of influence, it will be traveling at the exact same speed that it entered the sphere of influence in the first place, but, its direction will have been changed such that it is now heading toward our destination planet. This so called hyperbolic encounter, in addition to altering the direction of the probe, also alters the eccentricity and and energy of the probe's orbit with respect to the Sun. This is where the term slingshot came into use as the energy level of the probe is changed with respect to the Sun but not the passive turning planet. Another option in utilizing a planet to change direction is to add some velocity to the probe when it passes through its perigee altitude by means of another auxilliary rocket stage. Under this type of scenario, the exiting trajectory of the probe will be different from the entering trajectory.
The utilization of the swingby method to alter the trajectory of a probe can often be accomplished with less energy than the HTO method, though not always. The key advantage of the swingby approach is usually the reduction of time for the probe to reach its destination planet. The locations of all the planets concerned, and the control of the probe throughout, become much more critical for the success of such a mission.
The above discussion has addressed the how of delivering a space probe to the planet Pluto. The scenario is basically identical for sending probes to all the other planets with only the actual velocity numbers changing. For a more detailed discussion of how a rocket engine works and how the size of a rocket vehicle is determined, may I recommend you visit the Knowledge Database topic titled, "How Do Rockets Work?"
I hope this has been of some interest to you.
450 feet per second at a height of 6.5 feet
To estimate the time it would take for a space vehicle to travel from Earth to Neptune using the same speed, we need to know the approximate distance between Earth and Neptune. The average distance between Earth and Neptune is about 2.7 billion miles (4.35 billion kilometers).
If we assume that the space vehicle maintains the same speed throughout the entire journey, we can divide the distance to Neptune by the distance to the moon to find the approximate travel time:
Distance to Neptune / Distance to the Moon = Travel time to Neptune / Travel time to the Moon
(2.7 billion miles) / (238,855 miles) = Travel time to Neptune / 3 days
Solving for the travel time to Neptune, we get:
Travel time to Neptune = (2.7 billion miles / 238,855 miles) * 3 days
Travel time to Neptune ≈ 32,102 days
So, it would take approximately 32,102 days for the same space vehicle to reach Neptune if it maintained the same speed as it did when traveling to the moon.
To determine the time it would take for a space vehicle to travel from Earth to Neptune at the same speed, we need to gather some information and make a few calculations.
1. Calculate the average distance from Earth to the moon:
The average distance from Earth to the moon is approximately 384,400 kilometers.
2. Calculate the average distance from Earth to Neptune:
The average distance from Earth to Neptune is approximately 4.5 billion kilometers.
3. Calculate the time it takes to travel from Earth to the moon:
Given that it takes 3 days for a space vehicle to travel from Earth to the moon, we convert this time to hours:
3 days × 24 hours/day = 72 hours.
4. Calculate the average speed of the space vehicle:
To calculate average speed, we divide the distance traveled by the time taken:
Average speed = Distance / Time
Average speed to the moon = 384,400 km / 72 hours = 5,344.44 km/h
5. Calculate the time it would take to reach Neptune:
Now, we can use the average speed to calculate the time it would take to reach Neptune:
Time to reach Neptune = Distance to Neptune / Average speed
Time to reach Neptune = 4.5 billion km / 5,344.44 km/h
To simplify the calculation, we need to convert billion to million:
4.5 billion km = 4,500 million km.
Time to reach Neptune ≈ 4,500 million km / 5,344.44 km/h
This gives us an approximate time, in hours, for the vehicle to reach Neptune.