The gravitational force on a point mass m due to a point mass M is a gradient field with potential 

​U(r)equals=StartFraction GMm Over r EndFractionGMmr​, 

where G is the gravitational constant and 

requals=StartRoot x squared plus y squared plus z squared EndRootx2+y2+z2 

is the distance between the masses. What is a vector **t** that is always tangent to this​ shape?

A.left angle xy comma yz comma 2 xz right angle

left angle xy comma yz comma 2 xz right anglexy,yz,2xz

B.left angle x comma y comma z right angle

left angle x comma y comma z right anglex,y,z

C.left angle negative y comma z comma negative x right angle

left angle negative y comma z comma negative x right angle−y,z,−x

D.left angle negative yz comma negative xz comma 2 xy right angle

left angle negative yz comma negative xz comma 2 xy right angle

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!

The gravitational force on a point mass m due to a point mass M is a gradient field with potential U(r) = GMm/r, where G is the gravitational constant and r = sqrt(x^2 + y^2 + z^2) is the distance between the masses. What is a vector **t** that is always tangent to this shape? A. (xy, yz, 2xz) B. (x, y, z) C. (-y, z, -x) D. (-yz, -xz, 2xy)

Explore how to determine a vector tangent to the level surfaces of the gravitational potential field. Learn about gradient fields, gravitational potential, and vector calculus, with a step-by-step solution and formula breakdown. Suitable for advanced high school and college students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation