Moving objects using light

Question

Photon do not have mass but they have momentum, can we use laser to pick up golf ball and hurl it several yards away without burning it into crisp?

Answer 1

Hurl it several yards away

Rather than a simple yes/no, let's figure out some numbers - as usual I work with "round" numbers.

Mass of golf ball = 50 gram

"several yards" = 5 meters

Projectile launched at 45° with initial velocity $v$ travels for a time

$$t = 2\frac{v}{\sqrt{2}g}$$

during which time it will travel a distance

$$x = v_x t = \frac{v}{\sqrt{2}} \frac{v\sqrt{2}}{g}$$

So for a distance of 5 meters, we need an initial velocity of 7 m/s (again, round numbers).

Momentum $p = m\cdot v = 0.35 N\cdot s$.

Let us assume we use the most powerful laser on earth - this is at the NIF (National Ignition Facility - that word should give you a clue) and is capable of producing a frightening 500 TW of instantaneous power (about 2 MJ).

The momentum of light is given by

$$p = \frac{E}{c}$$

So the kick you get from this laser is $\frac{2\cdot 10^6}{3\cdot 10^8} = 0.007 N\cdot s$. If you reflect the light, you get a double kick (reverse the momentum) - leaving you still 25x short. You would need to build a more powerful laser with an energy of $50 MJ$.

So apart from the fact that there is no laser powerful enough on earth, what about the "burning" bit?

If you had a mirror capable of reflecting 99.999% of light, that one part in 10⁵ would leave you with 500 J absorbed in the surface layer. If the heat was distributed over the entire ball, it might survive that - 500 J over 50 gram will make the ball warm, but no worse than if you left it out in the sun on a Caribbean green.

However, since the pulse will be very fast, we can assume that the heat does not penetrate the ball but stays at the surface - let's say it stays within a 0.1 mm layer on the surface of the ball (this is after all supposed to be a reflector). The projected area of a golf ball (diameter a little over 40 mm) is about 12 cm² , giving us about 40 J per 10 mm³. The temperature of that layer will be significantly increased - enough to ablate the reflective coat. And as soon as the reflective (protective) coat is gone, the rest of the power of the laser can get to the heart of the ball.

So the answer is no, there is no laser powerful enough; and yes, if there was, it would burn to a crisp. Actually, a plasma ball more likely. Heck - you might get spontaneous fusion with that kind of heating. That's what the NIF is all about, after all...

Answer 2

I'm not sure about the “pick it up with a laser” part, but let's simplify by assuming that we can hit the golf ball from below the ground with photons. Let's further assume that the mass of the ball is $m_{ball} = 46\,\mathrm g$ and we want to hurl it five yards away ($\approx 4.6\,\mathrm m$). If the ground is level and we neglect air drag and wind, the best option is to fly in a parabola at an initial angle of 45 degrees. The range $R$ then depends on the initial velocity $v_i$ and the gravitational pull $g$ (see Wikipedia): $$ R = v_i^2 / g $$ With $R = 4.6\,\mathrm m$ and $g = 9.81\,\mathrm m\mathrm s^{-2}$ we get $v_i^2 \approx 45\, \mathrm m^2\mathrm s^{-2}$ and the impulse needed to hurl the ball as $$ p_{ball} = m_{ball} v_i \approx 0.3\,\mathrm {Ns} $$ If the ball's surface would be perfectly reflective (and assumed to be flat, i.e. a disc/cylinder, for simplification) each photon would be reflected and transmit twice its momentum $p_{ph}$. With this we can derive the number of photons $N_{ph}$ we need to hurl the ball: $$ p_{ball} = N_{ph} \times 2p_{ph} \approx 0.3\,\mathrm {Ns}\\ N_{ph} = 0.3\,\mathrm {Ns} / 2p_{ph} $$ The momentum of a photon and its energy are related by the speed of light $c$ $$ E_{ph} = p_{ph}c $$ so we can calculate the total energy $E$ of the photons: $$ E = N_{ph} \times E_{ph} = 0.3\,\mathrm {Ns} / 2p_{ph} \times p_{ph}c = 0.3\,\mathrm {Ns} \times c/2 \approx 45\,\mathrm {MJ} $$ As a step towards reality, let's assume that the ball is not perfectly reflective. So what would happen, if the ball would absorb, say, 10 % of the photons and heat up in the process. The energy absorbed would be in our simplistic calculation around $4.5\,\mathrm {MJ}$.
To put this in perspective, let's look at the vaporization of water (which has a very high heat capacity). To vaporize water from 25 °C we need 44 kJ/mol. So, we can vaporize about 100 moles of water with our absorbed energy, which equals 1.8 kg of water. Put this in the perspective of a ball weighing 46 g, I would guess that the ball would start to melt/burn/vaporize.

Answer 3

If the ball was completely reflective, yes. The concept was discussed e.g. here http://en.wikipedia.org/wiki/Solar_sail

edit: e.g.- radiation pressure is what keeps Sun from collapse