How much gravitational potential energy does a system comprising a 102-kg object and Earth have if the object is one Earth radius above the ground?
U = −3.19×109 J
How fast would a 102-kg object have to be moving at this height to have zero energy?

1)

2) 7909 m/s

Explanation:

1)

The potential energy of a system consisting of the Earth and an object in orbit around the Earth is given by

where

G is the gravitational constant

M is the Earth's mass

m is the mass of the object

r is the distance between the object and the Earth's centre

Here we have:

m = 102 kg is the mass of the object

r = 2R is the distance of the object from the Earth's centre, where

is the Earth's radius

Substituting,

2)

The total mechanical energy of the object is the sum of its potential energy and its kinetic energy:

where

E is the mechanical energy

U is the potential energy

K is the kinetic energy

Here we want the object to have zero energy, so

E = 0

This means that the kinetic energy is

The kinetic energy of the object can be rewritten as

where

m = 102 kg is the mass

v is the speed of the object

And solving for v, we find:

two charges are fixed in place on the x-axis as shown. the charge at x = 0 is negative and has magnitude q = 2 μcoulombs (1 μcoulomb = 10-6 coulombs). the charge at x = d, (d = +12 cm), is positive and has magnitude q = 4 μcoulombs. find a point x = xo along the x-axis at which the electric field is zero.