# Is It Possible To Move The Earth In A Direction? [duplicate]

According to Newton's 3rd law, when an object moves upward, the Earth moves by a very tiny factor in the opposite of the direction of the object. And when the object falls back, the earth moves back to its original position.

This lead me to think about rockets. When they leave the Earth, many of them don't come back. So, if we were to launch too many rockets with 1000x more speed from the position from which the earth would move towards the Sun (just an arbitrary direction); will it then be possible to move the Earth into the sun considering that none of the rockets which were launched came back?

## marked as duplicate by John Rennie, Kyle Kanos, Martin, Danu, Qmechanic♦Jan 8 '15 at 15:47

• In theory, yes. Are you asking if there is actually a chance of this happening? I doubt there's enough fuel on Earth, even assuming we hydrolyze all the water... – pentane Jan 8 '15 at 13:35
• Yea, I'm asking if there is actually a chance of this happening. – Always Learning Forever Jan 8 '15 at 13:37
• No. Definitely not. – Danu Jan 8 '15 at 13:41
• Possible duplicates: physics.stackexchange.com/q/29727/2451 , physics.stackexchange.com/q/38542/2451 , physics.stackexchange.com/q/56245/2451 and links therein. – Qmechanic Jan 8 '15 at 13:59
• A quick back of the envelope calculation: The energy required is about $\frac12 M_\oplus {V_\oplus}^2$, less a bit because we only need to cancel 99% of the Earth's orbital velocity. That's $2.5\times10^{33}$ joules. Compare that with the $4\times10^{22}$ joules in fossil fuel reserves, the $2.2\times10^{23}$ joules in uranium 238 deposits, the $5.5\times10^{24}$ joules of sunlight that strikes the face of Earth per year, or the $1.2\times10^{34}$ joules of sunlight the Sun produces per year. We would need to be able harness the entire output of the Sun. That's far, far future science fiction. – David Hammen Jan 8 '15 at 14:27

In order to show what would be needed to let Earth fall into the sun it would be easier to to view Earth as a rocket. When using this simplification I would be able to use the Tsiolkovsky rocket equation:

$$\Delta v = v_e \ln{\frac{m_0}{m_1}},$$

where $\Delta v$ is the change in velocity, $v_e$ the effective exhaust velocity, $m_0$ and $m_1$ the initial and final mass.

In case of letting the Earth fall into the sun, this would mean nearly zeroing out Earth's orbital velocity around the sun, which is roughly 29.78 km/s. But this value could be reduced a bit if you would only apply the change in velocity at aphelion and only such that the Earth and the sun would just "touch" each other (new perihelion would be equal to the sum of Earth's and sun's radii), namely this would require 27.29 km/s. For $m_0$ you would have to use the mass of the Earth, which is 5.97219$\times$10${}^{24}$ kg. In order to know how much mass would be needed to be expelled as fuel ($m_f = m_0 - m_1$) for a given exhaust velocity, you can rewrite the equation as follows,

$$m_f = m_0 \left(1 - \exp{\left(-\frac{\Delta v}{v_e}\right)}\right).$$

Values for $v_e$ will differ for different types of propulsion. For instance the highest known value achieved with a chemical reaction is lithium and fluorine which can achieve an exhaust velocity of 5320 m/s. However it might be more realistic to use a more common fuel, like hydrogen and oxygen (by electrolyzing water), which has an exhaust velocity of 4462 m/s. However both these exhaust velocities would require almost all the mass of the Earth, namely 5.93684$\times$10${}^{24}$ kg and 5.95901$\times$10${}^{24}$ kg or 99.4082% and 99.7793% respectively. The Earth however consists for 32.1% out of the element iron, so these kind of propellant methods will not be able to achieve this.

Another option might be nuclear propulsion, which has a specific impulses in the range of 6000 seconds, which is equivalent to an exhaust velocity of roughly 59 km/s. This would still require 2.21575$\times$10${}^{24}$ kg or 37.1012% of Earth's mass. But the Wikipedia article states that the theoretical maximum specific impulse would be 100,000 seconds, which is equivalent to an exhaust velocity of roughly 980 km/s. This would require 1.63848$\times$10${}^{23}$ kg or 2.74352% of Earth's mass. However the problem still will be the amount of available fuel.

I hope that this illustrates how hard this would be. I also simplified my approach, since you should also account for the potential energy to escape Earth's gravity well, how to transport the propellants out of Earth's atmosphere and our moon will also play a role.

Earth orbits the sun with a velocity of $29.78 km/s$. In order to fall into the sun that velocity would have to be reduced to approximately zero. The reason this velocity would have to be reduced to zero is an energy balance. Orbits are constantly balancing kinetic energy with gravitational potential energy, if we slow our velocity, we'll fall slightyl toward the sun picking up a bit of speed, that would let us orbit at our new distance from the sun. the most efficient orbital transfers apply two trusts one at the beginning to start the transfer to the new orbit, and one at the end to stabilize it. We wouldn't need the one at the end, but in order to get an orbital transfer that intersects with our sun our velocity would have to drop to near zero. So that's the $\Delta V$ for our rocket equation. If we used our best rockets we can get an exhaust velocity of about $4.4 km/s$. We would have to use 99.9% of earths mass as reaction fuel and the remaining 0.1% of earth would fall into the sun. If we were really dedicated to the cause we could try to use our most advanced ion thrusters with an exhaust velocity of up to around $210 km/s$. This would bring down our fuel mass to a more reasonable 14% of earth. However, the time-span required to consume 14% of earth one proton at a time would be enormous. But I suppose that's good, because the energy required to do so would take more than all of the current chemical energy on the planet, so we'd probably have to use solar power and just wait for more energy to arrive from the sun.

So in theory, even if we were trying to send earth into the sun, it would take a herculean effort and we wouldn't even make a dent by the time anyone reading this dies.

• Nice answer! Could you add sources for e.g. ion thruster exhaust velocity and current chemical energy on the planet? Also would you explain why the orbital velocity would have to be approximately zero in order for the Earth to fall into the Sun? – pentane Jan 8 '15 at 15:51
• @pentane I've added the thruster citation and tried to explain about orbital transfers. The chemical energy I didn't actually look, up but knew from the fact that the energy required would be comparable to most of the earth being comprised of chemically dense rocket fuels (from the rocket calculations) – Rick Jan 8 '15 at 16:20