# Velocity needed for bullet to circumnavigate earth and land at our feet

Lets say we fire a bullet from a very powerful gun at a height of 1 metre parallel to the surface of the Earth. Assume there is no air resistance and the only force acting on the bullet is from Earth's gravity. Also assume that the Earth is perfectly spherical and that nothing will get in the bullets way when it travels around the globe.

How fast must the bullet be fired such that is circumnavigates the Earth once and lands at our feet?

I know that its speed must be close to the speed needed for it to stay at an orbit height of $$1$$ metre ($$7925$$ m/s), but I am not sure how I can work the exact speed out.

I first thought the curvature of the earth wouldn't matter in the calculation and that I could just work out the answer for a flat distance of $$40 000$$ km (circumference of Earth), however I quickly realised that this would not work.

Is this situation I have described even possible? Would the orbit be circular or elliptic?

Any help with this question would be greatly appreciated.

• The orbit cannot be circular; if it were, the bullet would not land at your feet but rather at its original height. Jun 14 at 14:11
• I have a feeling this is not possible. A bounded Keplerian orbit is periodic, so it must pass through its initial point. The only way for the bullet to end at your feet is if it traveled in a straight-line path downward. It cannot circumnavigate the earth beforehand. Jun 14 at 14:14
• If a gun is pointed perpendicular to the surface of the earth then the bullet will go straight up. Did you mean to say that the gun is fired parallel to the surface of the earth? Jun 14 at 14:35
• This is Newton's cannonball: en.wikipedia.org/wiki/Newton%27s_cannonball . There is a simulator here that might produce the numerical value asked for: softpedia.com/get/Science-CAD/Tower-Simulation.shtml Searching Newtons cannonball finds many links. Jun 14 at 15:09
• How about firing the bullet essentially upwards at a velocity such that it comes down about 24 hours later and the Earth has rotated under it for a day, making for a circumnavigation after which it hits your foot? Jun 16 at 11:26

The situation as described is not possible. If Earth's gravity is the only force acting on the bullet, then the bullet's trajectory is Keplerian. That is, it is a conic section, of which the only bounded type is the ellipse. The Earth will be at one focus of the ellipse. If the bullet can circumnavigate the Earth, it will arrive back at its starting location, at the same height above Earth's surface. It is not possible to be at a lower elevation.

A related question is

What is the minimum speed required for the bullet, fired parallel to Earth's surface, to circumnavigate the Earth?

This speed corresponds to a situation in which the perigee of the orbit is $$R$$, Earth's radius. Note that this occurs $$180^\circ$$ from the launch point - it would land at the foot of someone on the other side of the Earth. In this case, the semi-major axis of the orbit is

$$a = \frac{1}{2} (R + h + R) = R + \frac{1}{2} h$$

where $$h$$ is the height at which the bullet is fired. By conservation of energy, we have

$$E_{\rm orb} = -\frac{K}{2 a} = -\frac{K}{2 R + h} = \frac{1}{2} v_0^2 - \frac{K}{R + h}$$

$$\implies v_0^2 = \frac{K}{R} \left[ \frac{2}{\left( 1 + \frac{h}{R} \right) \left( 2 + \frac{h}{R} \right)} \right]$$

where $$v_0$$ is the initial speed and $$K = G M$$ is Earth's gravitational parameter. Note that the speed required to maintain circular orbit satisfies $$v_c^2 = \frac{K}{R} \left[ \frac{1}{1 + \frac{h}{R}} \right]$$, which you already calculated. We have then that

$$v_0^2 = v_c^2 \left[ \frac{2}{2 + \frac{h}{R}} \right] \implies v_0 = v_c \sqrt{\frac{2}{2 + \frac{h}{R}}}$$

With $$h = 1 \text{ m}$$ and the Earth's radius approximately $$R = 6.38 \times 10^6 \text{ m}$$, the ratio $$\frac{h}{R} = 1.567 \times 10^{-7}$$ is extremely small. The factor $$\sqrt{\frac{2}{2 + \frac{h}{R}}} = 0.99999996$$ is practically unity, so the required speed is about $$39.2$$ parts per billion smaller than the speed required to maintain the circular orbit.

• The location of the bullet after a fixed flight time depends continuously on the initial speed. If $v_0=v_c$ than at the time when the bullet is half-way around the earth it will be 1m above ground. If $v_0=v_c*(1-39.2/billion)$ it will hit the ground and be at height 0. If $v_0$ is in between these two values it should be at some height smaller than 1m above ground then and will presumably hit the ground at some later time. Jun 15 at 6:17
• Thinking some more about this. At $v_c$ you have a circular orbit. For $v_0 < v_c$ you get an elliptic orbit where at $t=0$ you have the maximum distance to the center of the earth and the minimum speed of the orbit. Half way around the earth the bullet is at its minimal distance to the center of the earth and maximum speed. If it doesn't hit the ground at that point it will continue and come back exactly to your hand. /continued Jun 15 at 6:33
• In order to get continuity of the location depending on starting speed one would have to assume a perfect reflection when it hits the ground and then it would return exactly to you hand again. Jun 15 at 6:33

Others are mentioning the closed nature of Keplerian orbits, and that is accurate but misses that after one orbit of the bullet, we will be in a different position due to the Earth's rotation. So it is not forbidden that the bullet is at a different height when it passes us. (Though note that only particular combinations of our position and the firing direction will even make the bullet cross our position after orbiting. The simplest approach would be to fire east or west on the equator. You could also fire in any direction from a pole.)

However, assuming that

1. we're not allowed to change the properties of the Earth,
2. the bullet can't pass through the ground, and
3. our own height is much smaller than the radius of the Earth,

it is still not possible for the bullet to land at our feet.

The problem is this. If you fire the gun parallel to the Earth's surface, the firing point will be either the maximum radius (apocenter) or the minimum radius (pericenter) of the orbit, depending on whether the initial speed is slower or faster than the circular orbit speed. For the bullet to strike the ground, you need the firing point to be the apocenter. But the pericenter is achieved after exactly half an orbit, on the opposite side of the Earth after a time of about $$\pi R/\sqrt{GM/R}\simeq 42 \text{ minutes}$$ (where $$R$$ and $$M$$ are the radius and mass of the Earth and $$G$$ is the gravitational constant, so the circular orbit speed is $$\sqrt{GM/R}$$). After that time, the Earth's rotation has shifted our position by at most a fraction $$42\text{ minutes}/1\text{ day}\simeq 0.03$$ of the orbit.

If the pericenter is above the ground, the bullet will never land. If the pericenter is at ground level, the bullet lands at the pericenter, far away from us. If the pericenter is below the ground, the bullet lands even before pericenter.

However, it could be possible for the bullet to land at our feet after completing an orbit if any of the three assumptions above are relaxed.

1. If we make the Earth's rotation speed equal to or greater than the circular orbit speed at ground level, we could fire west and rotate east, meeting the bullet after it "circles the Earth" (or the Earth circles underneath it!) at whatever point we choose for it to land. Of course, the Earth would fly apart.
2. If the bullet can pass through the ground, we can let it go underground for most of its orbit and either pop up at our feet or cross our feet upon its second passage into the ground. Some care needs to be taken in this scenario, because an orbit inside an extended mass distribution is not Keplerian.
3. If we're allowed to be arbitrarily tall (in particular, much taller than the radius of the Earth), then we can make the orbital period arbitrarily long. With the right height, the time of pericenter can be 12 hours, just in time for the rotation of the Earth to carry us there.
• If you relax the initial direction constraint to allow other than parallel, you should also be able to find solutions.
– BCS
Jun 24 at 17:16

This is not possible if the Earth is spherical and there is no air resistance. All bounded Keplerian orbits are ellipses with one focus at the center of the Earth. If you require that the bullet pass through a point a few feet off the ground (where you fire the gun) and a point at your feet, these two points are collinear with the center of the Earth. And for any ellipse with a non-zero semiminor axis, the only way to have two points collinear with one of the foci is for the focus to lie between the points, not outside of the line segment connecting them as you envision.

The only way to get the bullet to pass through the desired point is to fire it straight up or straight down. In this case the orbit is an ellipse with a semi-minor axis of zero, and all pairs of points are collinear with the center of the Earth.

Frameshift:

The problem as stated is impossible for reasons that multiple authors have stated. Thus I will alter the problem a little bit:

The cannon is not quite level. Periapsis is zero, apoapsis is high enough to put it into an almost 24 hour orbit (I'm not going to take the time to work the math) such that when it reaches periapsis the Earth has rotated enough to put us where it lands. (Note, though, it's still going to be moving around 7,800m/s and the angle will be 0 degrees--it's going to bounce, not come to rest at our feet!)

As others have said, for the bullet to land on earth directly below the point from which it was shot, it cannot be in an orbit about the focus of an ellipse. Since $$s=\frac{1}{2}at^2$$, a body solely under gravitational acceleration of $$9.8\mathrm{m/sec^2}$$, whether dropped or projected horizontally from a height of $$1\mathrm{m}$$, will reach the ground—if it reaches it at all—in $$.45\mathrm{sec}$$. Supposing the shot fired due eastward , then, and (to avoid the Coriolis effect) at the earth’s equator, and taking earth’s circumference at the equator to be $$40,075,000\mathrm{m}$$, to touch ground at the shooter’s foot after traveling $$40,075,000\mathrm{m}$$ around earth, the bullet would need a horizontal velocity of $$89,055,556\mathrm{m/sec}$$. But this is well above the $$10,735\mathrm{m/sec}$$ escape velocity of a body moving tangentially eastward at earth’s equator. At that speed the bullet would follow a hyperbolic trajectory and never return to earth.

To arrive at the same place you have either a circle or an ellipse, in any case a closed path, so there is no possible speed with which you can achieve your purpose except just v=0 the ball drops to your foot.