55x times harder to get to the Sun?

[fizzy]

Hi,

there has been much media excitement about the launch of the Parker probe and some rather unhelpful attempts at dumbing down the language to the point where it no longer makes sense, or because science correspondence do not even understand the subject before they try to paraphrase PR material.

There have been various comments, generally seeming to come from a NASA press release about this mission needing 55 times more energy that a mission Mars.

Clearly the energy to escape the potential well of the Earths gravitational field does not depend upon whether you turn left or right at the lights after that.

Falling into the potential well created by 98% of the mass in the solar system would not seem to require too much energy on face value. Less in fact than trying to escape further out. It would seem that the idea is the need to kill tangential velocity in order to fall inwards rather than remaining in an earth-like orbit. They already exit Earth orbit "backwards" thus already being at less than orbital speed for this distance.

How does this end up needing 55x more energy than climbing the potential gradient to get to Mars ?

thanks.

 

[Andrew Mason]

Hi,

there has been much media excitement about the launch of the Parker probe and some rather unhelpful attempts at dumbing down the language to the point where it no longer makes sense, or because science correspondence do not even understand the subject before they try to paraphrase PR material.

There have been various comments, generally seeming to come from a NASA press release about this mission needing 55 times more energy that a mission Mars.

Clearly the energy to escape the potential well of the Earths gravitational field does not depend upon whether you turn left or right at the lights after that.

Falling into the potential well created by 98% of the mass in the solar system would not seem to require too much energy on face value. Less in fact than trying to escape further out. It would seem that the idea is the need to kill tangential velocity in order to fall inwards rather than remaining in an earth-like orbit. They already exit Earth orbit "backwards" thus already being at less than orbital speed for this distance.

How does this end up needing 55x more energy than climbing the potential gradient to get to Mars ?

thanks.

I expect it is involved in slowing down sufficiently to get close to the sun. The Earth is traveling at a tangential speed of 107,000 km/hr in its solar orbit. The escape velocity required to get to Mars is only about 40,000 km/hr. The kinetic energy of the probe itself at 107,000 km/hr is about 7 times greater at this speed. But since a rocket has to carry its fuel with it as it accelerates, it would need to carry much more than 7 times the fuel. The relationship between fuel and maximum speed is logarithmic. 55 times would seem about right.

AM

 

[Drakkith]

Clearly the energy to escape the potential well of the Earths gravitational field does not depend upon whether you turn left or right at the lights after that.

That's right. The direction mostly doesn't matter if you just want to escape Earth's sphere of influence.

Falling into the potential well created by 98% of the mass in the solar system would not seem to require too much energy on face value.

Oh boy does it ever! The problem is that to get extremely close to the Sun you need to kill off most of your orbital velocity. That's why the probe will launch on a trajectory that makes it leave the Earth while flying against our orbit around the Sun. We are traveling at about 29,000 m/s in our orbit, and cancelling enough of that takes a significant amount of fuel.

We commonly measure orbital maneuvers in terms of how much Δv (delta-v, which literally means change-in-velocity) they require. A mission to the Moon requires a delta-v of roughly 12,000 m/s. A mission to Mars requires a delta-v of about 15,000 m/s if you want to get into low orbit of Mars. However, just to intercept Mercury, without going an orbit (which would require additional delta-v) it takes 21,260 m/s delta-v. That is significantly more than it takes to get to Mars and we still aren't falling into the Sun.

The real problem is the fuel requirement, like Andrew pointed out above. As the required delta-v increases, the fuel requirements increases exponentially since you not only have to have fuel to accelerate the spacecraft up that higher velocity, you have to have fuel to accelerate that fuel as well! The Saturn V launch vehicle provided about 12,000 m/s delta-v to the Apollo spacecraft , just enough to get to the Moon. The first stage had a mass of 2.29 million kg, around 3/4 of the total mass of the launch vehicle, yet it only provided about 3,500 m/s delta-v, less than one third the total. Increasing the payload mass or the required delta-v greatly increases the amount of fuel needed.

Below is a link to a delta-v calculator so you can play with the numbers a little bit. Click the 'specific impulse' button and use 400 as the value. That's a realistic value for a hydrolox rocket engine. Then set your rocket's dry mass. I'd just set it to 1 to keep things simple. Now set your rocket's full mass to 2 and click 'calculate'. You should see a delta-v of about 2718. Now double the full mass to 4. Notice that the delta-v went up to 5437, double what we had before. So far it looks like a linear relationship, right? But let's keep going. Double the full mass to 8. Notice that the delta-v is only 8156, nowhere close to double what it was before. Double the mass to 16. The delta-v is now 10875. We just added 8 times more fuel than we had in our first calculation but we only gained about the same delta-v as our original 1 unit of fuel gave us. And it just keeps getting worse from there.

Delta-v calculator: http://www.strout.net/info/science/delta-v/

 

[fizzy]

thanks for the reply.

"The escape velocity required to get to Mars is only about 40,000 km/hr"

There is the escape velocity to get free of Earth's gravitational field, That does not depend on where you are going afterwards. Is the 40,000 km/hr the max speed needed to do the Mars trip in a humanly useful time?

I guess we need more details about the flight plan to understand it properly.Again it may be a question of how long. Just leaving the Earth "backwards" would mean < 107,000 km/hr and thus a decaying orbit but it could take years or centuries to spiral in. They are using a kind of negative slingshot off Venus to shed some speed too but I guess the fuel usage is just to get there by Christmas instead of taking five years.

The non linear amount of fuel needed is probably the key to the 55 times ( plus being in a hurry ) . Good point.

 

[Drakkith]

"The escape velocity required to get to Mars is only about 40,000 km/hr"

There is the escape velocity to get free of Earth's gravitational field, That does not depend on where you are going afterwards. Is the 40,000 km/hr the max speed needed to do the Mars trip in a humanly useful time?

I believe that's the speed to do what's called a Hohmann Transfer. It's a maneuver that requires the least amount of fuel and delta-v in most circumstances. You can do it for less if you have large bodies that you can swing around and use to do orbital braking or acceleration (gravitational slingshot) for you, but these require planets to be in specific positions and it may take much longer to get to your destination.

I guess we need more details about the flight plan to understand it properly.Again it may be a question of how long. Just leaving the Earth "backwards" would mean < 107,000 km/hr and thus a decaying orbit but it could take years or centuries to spiral in.

You would never spiral into the Sun. Ignoring gravitational influences from the other planets, you would just enter into a highly elliptical orbit and would stay there forever. Interactions with other planets, with the solar wind, and radiation pressure will all combine to alter your orbit over time in ways that are very difficult to predict in the long term though. You might end up getting shot out into the outer solar system by a close pass with Mercury or Venus, or they may swing you into an orbital path that takes you into the Sun itself.

They are using a kind of negative slingshot off Venus to shed some speed too but I guess the fuel usage is just to get there by Christmas instead of taking five years.

There will be seven close encounters with Venus over seven years, with each encounter designed to reduce the velocity of the spacecraft and send it into a lower orbit around the Sun. This saves a great deal of fuel, though it does take 7 years to get into the lowest orbit, whereas we could just do it immediately if we wanted to build a launch vehicle capable of providing the required delta-v. But that would probably be prohibitively expensive.

By the way, if you're interested in this topic at all, I highly recommend the game Kerbal Space Program. You will become intimately familiar with the basics of all of this very quickly.

 

[fizzy]

Delta-v calculator: http://www.strout.net/info/science/delta-v/

Thanks. I'm familiar with the dleta-v way of viewing manoeuvres , I think the key point that I was not accounting for is that it is not just getting to the Sun but getting there and establishing some kind of orbit. They could get there using less fuel, do a close pass without shedding all that speed but they would probably pass so fast that they would not get much data and would then be fired out of the SS entirely, or to return as a comet in 2,300 years time.

NASA says they will pass perihelion at 430,000 mph aka 688,000 km/h that is a lot of delta-v as you say.

So media reports that it takes a lot of fuel to get there are missing the point , it takes a lot to get whilst still traveling SLOWLY enough to do a useful mission.

 

[fizzy]

By the way, if you're interested in this topic at all, I highly recommend the game Kerbal Space Program.

Thanks, I may give that a look when I have some time. Sounds interesting and educational.

 

[Drakkith]

Thanks. I'm familiar with the dleta-v way of viewing manoeuvres , I think the key point that I was not accounting for is that it is not just getting to the Sun but getting there and establishing some kind of orbit. They could get there using less fuel, do a close pass without shedding all that speed but they would probably pass so fast that they would not get much data and would then be fired out of the SS entirely, or to return as a comet in 2,300 years time.

That's not how it works. The launch will immediately put the probe into an orbit with its aphelion at the same distance from the Sun as Earth is. Its perihelion will be inside the orbit of Mercury. This is a stable orbit and it requires no additional maneuvering to maintain it. The probe will not be launched into another orbit unless it interacts with another massive object like a planet. Spending less fuel would put its perihelion further away from the Sun, but it wouldn't change its aphelion. The orbit would be the same as it was before, just less eccentric (more circular shaped, less like on oval).

NASA says they will pass perihelion at 430,000 mph aka 688,000 km/h that is a lot of delta-v as you say.

The probe's velocity at perihelion is extremely fast, but that's okay. We don't actually need to counteract it. We would only need to do so if we want to reach a circular orbit at that distance.

So media reports that it takes a lot of fuel to get there are missing the point , it takes a lot to get whilst still traveling SLOWLY enough to do a useful mission.

No, the media is dead on. Just getting there takes a great amount of fuel. It would take several times that much if we wanted the probe to enter a circular orbit. Luckily the mission is designed around the probe making multiple close encounters with long periods further away from the Sun in between these encounters.

 

[russ_watters]

Again it may be a question of how long. Just leaving the Earth "backwards" would mean < 107,000 km/hr and thus a decaying orbit but it could take years or centuries to spiral in.

This is a common misconception, perhaps borne from pop media or sci fi: orbits do not decay on their own, they only decay if something is in the way, like an atmospere. If you start from any distance, with any non-zero tangential velocity - as long as it is enough to miss a collision- you are in a stable, elliptical orbit.

If they fired the engines enough to slow the probe so that it would miss the sun by 3 million miles, it would still loop back around to 90 million miles and they would still have a lot of energy to lose to get it into a stable, circular orbit.

 

[Andrew Mason]

thanks for the reply.

"The escape velocity required to get to Mars is only about 40,000 km/hr"

There is the escape velocity to get free of Earth's gravitational field, That does not depend on where you are going afterwards. Is the 40,000 km/hr the max speed needed to do the Mars trip in a humanly useful time?

40,000 km/hr is the velocity needed to escape Earth's gravity from a point on the Earth's surface. I used that only as a rough estimate for the speed to get to Mars.

I guess we need more details about the flight plan to understand it properly.Again it may be a question of how long. Just leaving the Earth "backwards" would mean < 107,000 km/hr and thus a decaying orbit but it could take years or centuries to spiral in. They are using a kind of negative slingshot off Venus to shed some speed too but I guess the fuel usage is just to get there by Christmas instead of taking five years.

Just to follow up on what Drakkith has said, if its speed tangential to the Earth orbit were to decrease by 107,000 km/hr the probe would plummet into the sun with an initial acceleration of about 2.3 m/sec^2. It would only be a matter of a couple of months before it would reach the sun surface. The plan is to lose a bit less than that and have enough tangential speed so that the probe misses the sun surface and goes around the sun (and then back out to where it started falling). The time to reach the closest distance to the sun would be about the same as a direct plummet into the sun.

AM

 

[swampwiz]

From the reference point of a satellite that it is at rest on a celestial body (i.e., theoretically defined as at rest with respect to the barycenter), the amount of energy to put that into a circular orbit (i.e., defined as being in free-fall but which stays at a constant distance from the barycenter) at the same distance from the barycenter is one-half the energy to put that into an escape trajectory - with the velocities relative to the barycenter as per the square-velocity rule for kinetic energy.

So a satellite that has just been given enough of a velocity change to be on an escape trajectory from Earth is in essence a satellite in orbit around the Sun along the Earth's orbit, and so the amount of velocity change to get on an escape velocity trajectory from the Sun would be ( 2^½ - 1 ), or about 2/5 of the orbital velocity of the Earth. However, to get to the Sun, the satellite would in essence have to fall into the Sun having no lateral velocity, so the the velocity change would be the orbital velocity of the Earth. Therefore, getting to the Sun would require the reciprocal of that expression, or about 5/2 the amount of velocity change to escape the Sun.

 

[James Demers]

It's a simple fact that de-orbiting involves a change in momentum, which requires (a) energy, and/or (b) interaction with another mass. Bouncing off of objects, like billiard balls do, gets messy at these velocities, so we use gravitation. You can get a whole lot of acceleration by "stealing" a tiny bit of a planet's momentum - the energy equivalelnt of hundreds of tons of fuel.
This can call for extreme measures: the Ulysses solar probe was sent all the way to Jupiter in order to get an orbit that passed over the sun's poles.

 

[Andrew Mason]

So a satellite that has just been given enough of a velocity change to be on an escape trajectory from Earth is in essence a satellite in orbit around the Sun along the Earth's orbit

?? Would its solar orbit not depend on the direction it was launched? How could it have a different speed than the Earth and still follow the Earth's orbit?

... so the amount of velocity change to get on an escape velocity trajectory from the Sun would be ( 2^½ - 1 ), or about 2/5 of the orbital velocity of the Earth. However, to get to the Sun, the satellite would in essence have to fall into the Sun having no lateral velocity, so the the velocity change would be the orbital velocity of the Earth. Therefore, getting to the Sun would require the reciprocal of that expression, or about 5/2 the amount of velocity change to escape the Sun.

In other words, it takes over 6 times as much energy to reach the sun than it would to escape from the sun. But, it appears to me, that your scenario of the satellite escaping earth, acquiring an Earth orbit and then plummeting to the sun would use a lot more energy than necessary. It would have the satellite initially escaping the Earth in the direction of the Earth's orbit and then later losing enough speed to enter a circular orbit of the same radius as the Earth's orbit around the sun (but far away from the Earth eg. on the opposite side of the sun from the Earth). There is no need to first escape the earth. Just launch it with 6 x the solar escape energy but pointed in the direction opposite to the Earth's orbit.

AM

 

[Buzz Bloom]

Today's APOD:

https://apod.nasa.gov/apod/ap180815.html .​

 

[Drakkith]

?? Would its solar orbit not depend on the direction it was launched? How could it have a different speed than the Earth and still follow the Earth's orbit?

Remember that Earth is pulling on the spacecraft as it moves away. If we launch a spacecraft with just barely enough velocity to escape into a solar orbit, then by the time it moves outside the Earth's sphere of influence it has been slowed by Earth's gravity to nearly a standstill compared to the Earth. So no matter what direction you launch the spacecraft , the resulting solar orbit is extremely close to that of Earth's.

Of course, the direction certainly matters if we give the spacecraft much more delta v than necessary to escape Earth. Launching retrograde will lower the spacecraft 's perihelion, while launching prograde will raise its aphelion. And obviously gravitational interactions with Earth and the other planets will perturb the spacecraft over time, changing it's orbit.

 

[Sean Bryan]

What about sending any craft out to Mars to use a gravity-assist flyby to redirect it to Sol? Would that also take 55x the energy?

 

[Drakkith]

What about sending any craft out to Mars to use a gravity-assist flyby to redirect it to Sol? Would that also take 55x the energy?

I'm not sure. Off the top of my head I want to say that it should take significantly less than 55x the energy. However, now you have to deal with the fact that your apoapsis (aphelion) is out near Mars. Not to mention the extra time spent in flight away from the Sun.

 

[Janus]

What about sending any craft out to Mars to use a gravity-assist flyby to redirect it to Sol? Would that also take 55x the energy?

It couldn't be done with a single fly-by. There are limits as to how much velocity one can gain/lose during a flyby. The Parker probe is scheduled to make 7 flybys of Venus during its mission.

 

[sophiecentaur]

What about sending any craft out to Mars to use a gravity-assist flyby to redirect it to Sol? Would that also take 55x the energy?

I think that they are far better off using Venus because it can give (after many fly-bys) a less eccentric orbit than you could get with Mars. If you think of a fly by as being a replacement for a rocket burn at the aphelion and that can't drop the orbit height at the point of the burn. See this link for a picture of the Parker orbits which always go round Venus. If you wanted to obtain an orbit within Venus's orbit, you would need a rocket burn at the perihelion and I guess that would involve too much fuel.
@Drakkith : A rather more long winded version of what you were saying.