Newton’s Universal Law of Gravity - Concluded

 

Mass and weight

 

You know the difference between mass and weight  your weight depends on where you are. Here’s why: We write Newton’s 2^nd law as  (but always keep in mind that it is more generally  ). The universal law of gravity, as applied to an object on the Earth’s surface, is . Here m_e and r_e indicate the mass and radius of Earth, and m is the mass of an object on the Earth. But F still equals ma, so  must be the acceleration m experiences. Insert the appropriate values: G = 6.67x10^-11, m_e = 5.97x10²⁴ and r_e = 6.38x10⁶, all in SI units. This provides an acceleration of 9.78 m/s². (The generally used value is, of course, 9.8). That’s why weight is different from mass  weight is tied to the planet you’re on.

 

Energy

 

We saw last semester that potential energy is related to force by . As always when dealing with potential energies, we need an arbitrary point of zero energy. By convention we take this at infinity. Multiplying by dr and integrating gives us . The left side integrates to U and the right side becomes  . (The gravitational force is directed towards Earth and the distance r is measured outward from Earth.) This is the gravitational energy of an object of mass m in the Earth’s gravitational field. Note that it is negative (which follows from our definition of zero potential energy at infinity). Our object’s kinetic energy is, of course, , so the total energy is

 

(1.1)                                                     

 

This is a useful tool: Remember conservation of energy? We lift an object of mass m a height h in Earth’s gravitational field and let it drop: The total energy at any point is  and this energy remains constant. The same applies here, except the potential energy is  Launch a projectile from the Earth’s surface (  ) to some height r, with an initial velocity v. Conservation of energy requires

 

(1.2)                                           

(Notice the mass of the projectile cancels out  this works for any object). If we look for the height where the object stops, the kinetic energy term on the right becomes zero. If we want to find what initial velocity is needed to completely escape Earth’s gravity, set the entire right-hand side to zero. Then solve for v_i to get the “escape velocity”. The result is (as you should confirm) . Inserting values provides 1.12x10⁴ m/s (about 25,000 mph). You might calculate the escape velocity of Phobos, the larger of the moons of Mars. Its mass is 1.08x10¹⁶ kg and its diameter is 22.2 km. It could be tricky walking on Phobos.