I was wondering whether it would be possible for a bullet shot from gun to become moon's satellite. How fast the bullet would have to be?

2 Answers

pohnpei397's profile pic

pohnpei397 | College Teacher | (Level 3) Distinguished Educator

Posted on

I assume that you are talking about shooting a bullet from the Earth and having it become a satellite of the moon.  I do not know if it would be possible, but here are some thoughts:

  • It's easy enough to figure out how fast the bullet needs to be going to get out of Earth's gravity and not fall back to Earth.  The equation for that is

Ve (escape velocity) = square root of (2Gm/r) where G is the universal gravitational constant, m is the mass of Earth and r is the radius from the Earth's center to the place from which you are shooting the bullet.

  • The major reason I think this would be difficult is because I doubt the bullet could be made to go at the exact velocity needed to get it to enter orbit around the moon (not so slow that it falls to the moon, not so fast that it can not be caught by the moon's gravitational field).
neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

Anything projected from the surface of a planet should work aginst the gravity to go up or away from the planet. If  v is  the velocity of bullet of mass m, projected on a surface of the planet (here, moon.), of mass M and radius R, then the object's (here bullet's)  kinetic energy and potential energy should exceed that of its potential energy. The kinetic and poential eenergy of the object  are given by:

KE = mv^2/2 and

PE = m*g, where g is the acceleration due to gravitational attraction, given by:

mg =  GMm/R^2. Here, G is the unineversal gravitational constant = 6.673*10-11 Nm/kg^2, by tables. Or

g = GM/R^2

To escape the moon's gravity, the object's (bullet' s) kinetic energy should be more than  of it's surfacial potential energy, for which the required condition is:

bullet's KE > = bullet's  PE  on the surface of the planet (moon) or

mv^2/2 >= mg, or

V^2  =  2g  or

V^2 > = 2 moon's surfacial acceleration due to moon's gravity.

= 2* (GM/R^2), or

v = (2GM)^(1/2) / R.

For the moon of us (earth), by tables, M = 7.353*10^22 kg, R = 1.738* 10^6 meters. Therefore the escape velocity for the bullet (or any other thing projected from the moon's surface) is given by:

v = [(2*6.672*10^-11 Nm/kg^2) * (7.353*10^22 kg)]^(1/2) /(1738*10^6 m)

=1.274509489 meter/ second.