Find speed of 1 kg object when it reaches close to Earth's surface from a long distance after it is released from rest as shown in the diagram. [Given \( R_e = 6400 \, \text{km}, g_s = 9.8 \, \text{m/s}^2 \)]
Show Hint
The speed of an object falling towards Earth can be found using conservation of energy. The potential energy lost is equal to the kinetic energy gained, and the speed at Earth's surface can be derived from this relationship.
Step 1: Understanding the problem.
We are asked to find the speed of a 1 kg object when it reaches close to the Earth's surface from a long distance after being released from rest. We can use the concept of conservation of mechanical energy, where the potential energy lost by the object is converted into kinetic energy.
Step 2: Applying the conservation of energy.
The total mechanical energy at any point in the object's path is conserved. The initial potential energy of the object at a far distance is converted into kinetic energy as the object approaches the Earth's surface. The potential energy at a distance far from Earth is given by:
\[
U = - \frac{GMm}{r}
\]
Where:
- \( G \) is the gravitational constant (\( 6.674 \times 10^{-11} \, \text{N} \cdot \text{m}^2/\text{kg}^2 \)),
- \( M \) is the mass of the Earth (\( 5.97 \times 10^{24} \, \text{kg} \)),
- \( m \) is the mass of the object (\( 1 \, \text{kg} \)),
- \( r \) is the distance from the center of the Earth.
At a large distance, the object is released from rest, so its initial kinetic energy is zero. As it falls towards Earth, its potential energy is converted into kinetic energy.
\[
K = \frac{1}{2} m v^2
\]
Step 3: Solving for the speed at Earth's surface.
When the object reaches close to the Earth's surface, the potential energy becomes:
\[
U_{\text{surface}} = - \frac{GMm}{R_e}
\]
Where \( R_e \) is the radius of the Earth (\( 6400 \, \text{km} \)).
The total energy at the beginning is equal to the total energy at the Earth's surface:
\[
K + U_{\text{surface}} = 0
\]
\[
\frac{1}{2} m v^2 - \frac{GMm}{R_e} = 0
\]
Solving for \( v \), the speed at the Earth's surface:
\[
v = \sqrt{\frac{2GM}{R_e}}
\]
Substituting the known values:
\[
v = \sqrt{\frac{2 \times 6.674 \times 10^{-11} \times 5.97 \times 10^{24}}{6400 \times 10^3}}
\]
\[
v \approx 11.2 \, \text{km/s}
\]
Step 4: Conclusion.
Therefore, the speed of the object when it reaches close to Earth's surface is approximately \( 11.2 \, \text{km/s} \).
Final Answer: 11.2 km/s
