A body of weight 72 N moves from the surface of earth at a height half of the radius of earth, then gravitational force exerted on it will be

Question

Two identical particles each of mass \( m \) go around a circle of radius \( a \) under the action of their mutual gravitational attraction. The angular speed of each particle will be:

Answer 1

To solve this problem, we need to understand how the gravitational force changes as the distance from the center of the Earth increases. The gravitational force acting on an object is given by Newton's law of universal gravitation:

F = \frac{{G \cdot m_1 \cdot m_2}}{{r^2}}

where:

F is the gravitational force between the two objects.
G is the gravitational constant.
m_1 and m_2 are the masses of the two objects.
r is the distance between the centers of the two objects.

For an object on the surface of the Earth, the gravitational force is also expressed as F = mg, where m is the mass of the object and g is the acceleration due to gravity at the surface of the Earth. In this case, the weight of the object is 72 N.

When the object is moved to a height above the Earth's surface, the distance from the center of the Earth becomes R + h, where R is the radius of the Earth and h is the height above the Earth's surface. In this problem, h = \frac{R}{2}, so the distance from the center is 1.5R.

The gravitational force at a height h is given by:

F_h = \frac{{G \cdot m_1 \cdot m_2}}{{(R + h)^2}} = \frac{{G \cdot m_1 \cdot m_2}}{{(1.5R)^2}}

Dividing this by the force at the surface:

\frac{{F_h}}{{F}} = \frac{{R^2}}{{(1.5R)^2}} = \frac{1}{(1.5)^2} = \frac{1}{2.25}

Plugging in the weight of the object:

F_h = 72 \times \frac{1}{2.25} = 72 \times \frac{4}{9} = 32 \text{ N}

Thus, the gravitational force exerted on the object when it is at a height \frac{R}{2} above the Earth's surface is 32 N.

Answer 2