Home
Class 11
PHYSICS
The gravitational potential due to a mas...

The gravitational potential due to a mass distribution is `V = 3X^(2) Y+ Y^(3) Z`. Find the gravitational filed.

Text Solution

AI Generated Solution

The correct Answer is:
To find the gravitational field from the given gravitational potential \( V = 3X^2 Y + Y^3 Z \), we will follow these steps: ### Step 1: Understand the relationship between gravitational field and potential The gravitational field \( \mathbf{E} \) is related to the gravitational potential \( V \) by the equation: \[ \mathbf{E} = -\nabla V \] where \( \nabla V \) is the gradient of the potential. ### Step 2: Write the gradient in three dimensions In three-dimensional Cartesian coordinates, the gradient of a scalar function \( V(x, y, z) \) is given by: \[ \nabla V = \left( \frac{\partial V}{\partial x} \hat{i} + \frac{\partial V}{\partial y} \hat{j} + \frac{\partial V}{\partial z} \hat{k} \right) \] Thus, the gravitational field can be expressed as: \[ \mathbf{E} = -\left( \frac{\partial V}{\partial x} \hat{i} + \frac{\partial V}{\partial y} \hat{j} + \frac{\partial V}{\partial z} \hat{k} \right) \] ### Step 3: Calculate the partial derivatives of \( V \) Given \( V = 3X^2 Y + Y^3 Z \), we will compute the partial derivatives: 1. **Partial derivative with respect to \( x \)**: \[ \frac{\partial V}{\partial x} = \frac{\partial}{\partial x}(3X^2 Y + Y^3 Z) = 6XY \] 2. **Partial derivative with respect to \( y \)**: \[ \frac{\partial V}{\partial y} = \frac{\partial}{\partial y}(3X^2 Y + Y^3 Z) = 3X^2 + 3Y^2 Z \] 3. **Partial derivative with respect to \( z \)**: \[ \frac{\partial V}{\partial z} = \frac{\partial}{\partial z}(3X^2 Y + Y^3 Z) = Y^3 \] ### Step 4: Substitute the derivatives into the field equation Now we can substitute these derivatives back into the equation for the gravitational field: \[ \mathbf{E} = -\left( 6XY \hat{i} + (3X^2 + 3Y^2 Z) \hat{j} + Y^3 \hat{k} \right) \] ### Step 5: Write the final expression for the gravitational field Thus, the gravitational field \( \mathbf{E} \) is: \[ \mathbf{E} = -6XY \hat{i} - (3X^2 + 3Y^2 Z) \hat{j} - Y^3 \hat{k} \] ### Final Answer: \[ \mathbf{E} = -6XY \hat{i} - (3X^2 + 3Y^2 Z) \hat{j} - Y^3 \hat{k} \] ---

To find the gravitational field from the given gravitational potential \( V = 3X^2 Y + Y^3 Z \), we will follow these steps: ### Step 1: Understand the relationship between gravitational field and potential The gravitational field \( \mathbf{E} \) is related to the gravitational potential \( V \) by the equation: \[ \mathbf{E} = -\nabla V \] where \( \nabla V \) is the gradient of the potential. ...
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • GRAVITATION

    DC PANDEY ENGLISH|Exercise Exercise 13.5|6 Videos
  • GRAVITATION

    DC PANDEY ENGLISH|Exercise Exercise 13.6|3 Videos
  • GRAVITATION

    DC PANDEY ENGLISH|Exercise Exercise 13.3|5 Videos
  • GENERAL PHYSICS

    DC PANDEY ENGLISH|Exercise INTEGER_TYPE|2 Videos
  • KINEMATICS

    DC PANDEY ENGLISH|Exercise INTEGER_TYPE|10 Videos

Similar Questions

Explore conceptually related problems

The gravitational potential due to a mass distribution is V=A/sqrt(x^2+a^2) . Find the gravitational field.

The gravitational potential due to a mass distrubution is given by V=(8x)/(x^(2)+p^(2)) . Find gravitational field at x = p.

The gravitational potential in a region by V = (20x + 40y)j/kg. Find out the gravitational field (in newton /kg) at a point co-ordinates (2,4). Also find out the magnitude of the gravitational force on a particle of 0.250 kg placed at the point (2,4) .

On the x - axis and at a distance x from the origin, the gravitational field due to a mass distribution is given by (Ax)/((x^2+a^2)^(3//2)) in the x - direction. The magnitude of gravitational potential on the x - axis at a distance x, taking its value to be zero at infinity , is :

The gravitational potential due to earth at infinite distance from it is zero. Let the gravitational potential at a point P be -5 J kg^(-1) . Suppose, we arbitrarily assume the gravitational potential at infinity to be +10 J kg^(-1) , then the gravitational potential at P will be

The gravitational field due to a mass distribution is E=K//x^(3) in the x - direction. ( K is a constant). Taking the gravitational potential to be zero at infinity, its value at a distance x is

The gravitational field due to a mass distribution is E=(A)/(x^(2)) in x-direction. Here, A is a constant, Taking the gravitational potential to be zero at infinity, potential at x is

Gravitational potential at a distance 'r' from a point mass 'm' is V = -(GM)/(r) Find gravitational field strength at that point.

Gravitational potential in x-y plan varies with x and y coordinates as V = x^(2)y+2xy Find gravitational field strength E .

If gravitational potential is V = xy^(2) , find the gravitational field at (2,1)