To solve this problem, we need to determine a vector t that is always tangent to the level surfaces of the potential function $U(r) = \frac{GMm}{r}$, where $r = \sqrt{x^2 + y^2 + z^2}$.

Step 1: Understanding the gradient and tangent vectors

The gradient of a scalar field $U(r)$ points in the direction of the greatest increase of the function and is perpendicular to the level surfaces of the function. For a vector to be tangent to the level surfaces, it must be perpendicular to the gradient of the potential $U(r)$.

Step 2: Compute the gradient of $U(r)$

The potential function $U(r) = \frac{GMm}{r}$ depends on the distance $r = \sqrt{x^2 + y^2 + z^2}$, so we first compute the gradient of $r$.

The gradient of $r$ is:

$\nabla r = \left( \frac{\partial r}{\partial x}, \frac{\partial r}{\partial y}, \frac{\partial r}{\partial z} \right) = \left( \frac{x}{r}, \frac{y}{r}, \frac{z}{r} \right)$

Thus, the gradient of $U(r)$ is:

$\nabla U(r) = \frac{dU}{dr} \nabla r = -\frac{GMm}{r^2} \left( \frac{x}{r}, \frac{y}{r}, \frac{z}{r} \right) = -\frac{GMm}{r^3} \left( x, y, z \right)$

Step 3: Identify the tangent vector

To find a vector that is tangent to the level surface, it must be perpendicular to the gradient $\nabla U(r)$. This means the dot product between the tangent vector and the gradient should be zero.

The gradient $\nabla U(r)$ is proportional to $\left( x, y, z \right)$, so the tangent vector must satisfy:

$x t_x + y t_y + z t_z = 0$

Step 4: Test the given options

Now, let's check which of the given vector options satisfies this condition:

• Option A: $\left( xy, yz, 2xz \right)$ $x(xy) + y(yz) + z(2xz) = x^2 y + y^2 z + 2x z^2$ This is not necessarily zero, so this option is incorrect.

• Option B: $\left( x, y, z \right)$ $x(x) + y(y) + z(z) = x^2 + y^2 + z^2$ This is not zero, so this option is incorrect.

• Option C: $\left( -y, z, -x \right)$ $x(-y) + y(z) + z(-x) = -xy + yz - zx = 0$ This is always zero, so this option is correct.

• Option D: $\left( -yz, -xz, 2xy \right)$ $x(-yz) + y(-xz) + z(2xy) = -xyz - xyz + 2xyz = 0$ This simplifies to zero, but the form is not as generally correct as Option C.

Conclusion:

The vector t that is always tangent to the level surfaces of the potential is given by:

C. $\left( -y, z, -x \right)$

Would you like a further breakdown of the steps?

Here are 5 related questions:

1. What is the physical meaning of the gradient of a scalar field?
2. How do you calculate the gradient of a potential function?
3. What is the geometric significance of a tangent vector in a level surface?
4. Why must the dot product of a gradient and a tangent vector be zero?
5. How does the gravitational potential change as the distance between two masses increases?

Tip: The dot product is a key tool for finding perpendicular vectors!