utron star has a mass of 2.0 xX kg (about the mass of our sun) and a P ountain). Suppose an object falls from rest near the surface of such a star. H stance of 0.015 m? (Assume that the gravitational force is constant over the IM
Gravitational force
In nature, every object is attracted by every other object. This phenomenon is called gravity. The force associated with gravity is called gravitational force. The gravitational force is the weakest force that exists in nature. The gravitational force is always attractive.
Acceleration Due to Gravity
In fundamental physics, gravity or gravitational force is the universal attractive force acting between all the matters that exist or exhibit. It is the weakest known force. Therefore no internal changes in an object occurs due to this force. On the other hand, it has control over the trajectories of bodies in the solar system and in the universe due to its vast scope and universal action. The free fall of objects on Earth and the motions of celestial bodies, according to Newton, are both determined by the same force. It was Newton who put forward that the moon is held by a strong attractive force exerted by the Earth which makes it revolve in a straight line. He was sure that this force is similar to the downward force which Earth exerts on all the objects on it.
![### Educational Content
**Problem Statement:**
A neutron star has a mass of \(2.0 \times 10^{30}\) kg (about the mass of our sun) and a radius of \(5.0 \times 10^3\) m (about the height of a good-sized mountain). Suppose an object falls from rest near the surface of such a star. How fast would this object be moving after it had fallen a distance of 0.015 m? (Assume that the gravitational force is constant over the distance of the fall and that the star is not rotating.)
**Given Data:**
- Mass of the neutron star, \( M = 2.0 \times 10^{30} \) kg
- Radius of the neutron star, \( R = 5.0 \times 10^3 \) m
- Distance fallen by the object, \( d = 0.015 \) m
**Equation for Velocity:**
The velocity \( v \) of an object falling under gravity can be calculated using the equation for gravitational potential energy and kinetic energy.
1. Gravitational force \( F \) is \( F = \frac{GMm}{R^2} \), where \( G \) is the gravitational constant (\(6.674 \times 10^{-11} \, \text{N} \cdot \text{(m/kg)}^2 \)).
2. Potential energy \( U \) is given by \( U = \frac{GMm}{R} \).
3. The kinetic energy \( K \) will be \( K = \frac{1}{2}mv^2 \).
Since the object falls from rest:
\[ \Delta U = \frac{GMm}{R} - \frac{GMm}{R + d} \]
\[ \Delta U = \frac{GMm}{R^2} \cdot d \]
Since the initial kinetic energy is zero and all the potential energy converts into kinetic energy:
\[ \frac{1}{2}mv^2 = \frac{GMm}{R^2} \cdot d \]
\[ v = \sqrt{\frac{2GM \cdot d}{R^2}} \]
Users are encouraged to solve for \( v \) by inserting the given values:
- \( G = 6.674 \times 10^{-11} \, \text](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3f0a09bd-ef71-43b6-abcd-653755da1dfb%2Fa4816484-0f15-47d8-bcd6-cd8bec18dd95%2F4hz1gs_processed.png&w=3840&q=75)
Trending now
This is a popular solution!
Step by step
Solved in 3 steps with 3 images
