A projectile is launched from Earth’s surface and reaches a maximum height of h. Assume that h << RE, where RE is the radius of the Earth.
Show that the variation in gravitational potential energy of Earth with height can be written as U=mgz(1-z/r), where g=GM/R^2and z is the height of an object above the Earth’s surface. Hint, recall that you can set the zero-point of the potential energy to any convenient value.
To derive the expression for the variation in gravitational potential energy with height, we need to consider the gravitational force acting on an object at a height z above the Earth's surface.
1. Start with the expression for gravitational potential energy (U) of an object close to the Earth's surface:
U = -GMm/r
where G is the gravitational constant, M is the mass of the Earth, m is the mass of the object, and r is the distance between the object and the center of the Earth.
2. Assume that the height h is much smaller than the radius of the Earth (h << RE). We can then approximate the distance r as:
r ≈ RE
where RE is the radius of the Earth.
3. Using the approximation r ≈ RE, we can rewrite the expression for gravitational potential energy as:
U ≈ -GMm/RE
4. Now, let's consider the height z of the object above the Earth's surface. We can express this height as the difference between the radius of the Earth (RE) and the distance from the center of the Earth to the object (r):
z = RE - r
5. Rearrange the equation to solve for r:
r = RE - z
6. Substitute this value of r into the expression for U:
U ≈ -GMm/(RE - z)
7. We can factor out a negative sign and rearrange the equation to get:
U ≈ GMm/(z - RE)
8. Recall that the acceleration due to gravity (g) near the Earth's surface is given by:
g = GM/RE²
9. Substitute g into the equation:
U ≈ mg/(z - RE)
10. Finally, we can rewrite the expression for U by multiplying the numerator and denominator by -1 to match the desired form:
U ≈ mg(z - RE)/-(z - RE)
11. Simplifying the expression further, we get:
U ≈ mgz/(RE - z)
12. Rearranging this expression, we finally obtain the desired form:
U ≈ mgz(1 - z/RE)
Therefore, the variation in gravitational potential energy of Earth with height is given by the expression U = mgz(1 - z/RE), where g = GM/RE² and z is the height of the object above Earth's surface.
Note: The zero-point of the potential energy can be set to any convenient value, and in this derivation, we assumed it to be at the surface of the Earth.