Answer :
### Understanding the Value of g
The value of [tex]\( g \)[/tex] is [tex]\( -9.8 \, \text{m/s}^2 \)[/tex]. This constant represents the acceleration due to gravity near the Earth's surface.
1. Acceleration Due to Gravity: The value indicates that any object in free fall near the Earth's surface will accelerate downwards at a rate of [tex]\( 9.8 \, \text{m/s}^2 \)[/tex].
2. Direction of Gravitational Force: The negative sign signifies that the force of gravity acts in the direction towards the center of the Earth (downwards).
### Numerical Problems Solved
#### Problem 1: Gravitational Force Between the Sun and the Earth
- Given:
- Mass of the Sun ([tex]\( M_s \)[/tex]): [tex]\( 2 \times 10^{30} \, \text{kg} \)[/tex]
- Mass of the Earth ([tex]\( M_e \)[/tex]): [tex]\( 6 \times 10^{24} \, \text{kg} \)[/tex]
- Distance between them ([tex]\( d \)[/tex]): [tex]\( 1.5 \times 10^8 \, \text{km} \)[/tex] = [tex]\( 1.5 \times 10^{11} \, \text{m} \)[/tex]
- Formula:
[tex]\[ F = \frac{G M_s M_e}{d^2} \][/tex]
- Calculation:
Plugging in the values (with [tex]\( G = 6.67430 \times 10^{-11} \, \text{m}^3 \text{kg}^{-1} \text{s}^{-2} \)[/tex]):
[tex]\[ F = \frac{6.67430 \times 10^{-11} \times 2 \times 10^{30} \times 6 \times 10^{24}}{(1.5 \times 10^{11})^2} \][/tex]
This calculates to:
[tex]\[ F \approx 3.559626666666666e^{22} \, \text{N} \][/tex]
#### Problem 2: Mass of an Object from Gravitational Force
- Given:
- Gravitational Force ([tex]\( F \)[/tex]): [tex]\( 1.334 \times 10^{-3} \, \text{N} \)[/tex]
- Mass of one object ([tex]\( m_1 \)[/tex]): [tex]\( 200 \, \text{kg} \)[/tex]
- Distance between objects ([tex]\( d \)[/tex]): [tex]\( 10 \, \text{m} \)[/tex]
- Formula:
[tex]\[ m_2 = \frac{F d^2}{G m_1} \][/tex]
- Calculation:
Plugging in the values (with [tex]\( G = 6.67430 \times 10^{-11} \)[/tex]):
[tex]\[ m_2 = \frac{1.334 \times 10^{-3} \times 10^2}{6.67430 \times 10^{-11} \times 200} \][/tex]
This calculates to:
[tex]\[ m_2 \approx 9993557.37680356 \, \text{kg} \][/tex]
#### Problem 3: Mass of a Person on Earth
- Given:
- Weight of the person ([tex]\( W \)[/tex]): [tex]\( 977 \, \text{N} \)[/tex]
- Mass of the Earth ([tex]\( M_e \)[/tex]): [tex]\( 6 \times 10^{24} \, \text{kg} \)[/tex]
- Radius of the Earth ([tex]\( r \)[/tex]): [tex]\( 6400 \, \text{km} \)[/tex] = [tex]\( 6400 \times 10^3 \, \text{m} \)[/tex]
- Formula:
[tex]\[ W = G \frac{M_e m}{r^2} \][/tex]
Solving for [tex]\( m \)[/tex]:
[tex]\[ m = \frac{W r^2}{G M_e} \][/tex]
- Calculation:
Plugging in the values (with [tex]\( G = 6.67430 \times 10^{-11} \)[/tex]):
[tex]\[ m = \frac{977 \times (6400 \times 10^3)^2}{6.67430 \times 10^{-11} \times 6 \times 10^{24}} \][/tex]
This calculates to:
[tex]\[ m \approx 99.93037971522608 \, \text{kg} \][/tex]
#### Problem 4: Height from the Earth's Surface for Changed Weight
- Given:
- Weight at earth's surface ([tex]\( W_1 \)[/tex]): [tex]\( 300 \, \text{N} \)[/tex]
- Weight at height [tex]\( h \)[/tex] ([tex]\( W_2 \)[/tex]): [tex]\( 200 \, \text{N} \)[/tex]
- Radius of the Earth ([tex]\( r_e \)[/tex]): [tex]\( 6380 \, \text{km} \)[/tex] = [tex]\( 6380 \times 10^3 \, \text{m} \)[/tex]
- Formula:
Combine the weight formulas:
[tex]\[ W_1 = G \frac{M_e m}{r_e^2} \][/tex]
and
[tex]\[ W_2 = G \frac{M_e m}{(r_e + h)^2} \][/tex]
We get:
[tex]\[ \frac{W_2}{W_1} = \frac{r_e^2}{(r_e + h)^2} \][/tex]
Solving for [tex]\( h \)[/tex]:
[tex]\[ \frac{200}{300} = \frac{r_e^2}{(r_e + h)^2} \][/tex]
[tex]\[ \frac{2}{3} = \frac{(6380 \times 10^3)^2}{(6380 \times 10^3 + h)^2} \][/tex]
- Calculation:
Simplifying the equation, we get:
[tex]\[ (r_e + h)^2 = \frac{3}{2} r_e^2 \][/tex]
Solving for [tex]\( h \)[/tex]:
[tex]\[ h = \sqrt{\frac{3}{2} r_e^2} - r_e \][/tex]
This calculates to:
[tex]\[ h \approx 521216582732.1647 \, \text{m} \][/tex]
The value of [tex]\( g \)[/tex] is [tex]\( -9.8 \, \text{m/s}^2 \)[/tex]. This constant represents the acceleration due to gravity near the Earth's surface.
1. Acceleration Due to Gravity: The value indicates that any object in free fall near the Earth's surface will accelerate downwards at a rate of [tex]\( 9.8 \, \text{m/s}^2 \)[/tex].
2. Direction of Gravitational Force: The negative sign signifies that the force of gravity acts in the direction towards the center of the Earth (downwards).
### Numerical Problems Solved
#### Problem 1: Gravitational Force Between the Sun and the Earth
- Given:
- Mass of the Sun ([tex]\( M_s \)[/tex]): [tex]\( 2 \times 10^{30} \, \text{kg} \)[/tex]
- Mass of the Earth ([tex]\( M_e \)[/tex]): [tex]\( 6 \times 10^{24} \, \text{kg} \)[/tex]
- Distance between them ([tex]\( d \)[/tex]): [tex]\( 1.5 \times 10^8 \, \text{km} \)[/tex] = [tex]\( 1.5 \times 10^{11} \, \text{m} \)[/tex]
- Formula:
[tex]\[ F = \frac{G M_s M_e}{d^2} \][/tex]
- Calculation:
Plugging in the values (with [tex]\( G = 6.67430 \times 10^{-11} \, \text{m}^3 \text{kg}^{-1} \text{s}^{-2} \)[/tex]):
[tex]\[ F = \frac{6.67430 \times 10^{-11} \times 2 \times 10^{30} \times 6 \times 10^{24}}{(1.5 \times 10^{11})^2} \][/tex]
This calculates to:
[tex]\[ F \approx 3.559626666666666e^{22} \, \text{N} \][/tex]
#### Problem 2: Mass of an Object from Gravitational Force
- Given:
- Gravitational Force ([tex]\( F \)[/tex]): [tex]\( 1.334 \times 10^{-3} \, \text{N} \)[/tex]
- Mass of one object ([tex]\( m_1 \)[/tex]): [tex]\( 200 \, \text{kg} \)[/tex]
- Distance between objects ([tex]\( d \)[/tex]): [tex]\( 10 \, \text{m} \)[/tex]
- Formula:
[tex]\[ m_2 = \frac{F d^2}{G m_1} \][/tex]
- Calculation:
Plugging in the values (with [tex]\( G = 6.67430 \times 10^{-11} \)[/tex]):
[tex]\[ m_2 = \frac{1.334 \times 10^{-3} \times 10^2}{6.67430 \times 10^{-11} \times 200} \][/tex]
This calculates to:
[tex]\[ m_2 \approx 9993557.37680356 \, \text{kg} \][/tex]
#### Problem 3: Mass of a Person on Earth
- Given:
- Weight of the person ([tex]\( W \)[/tex]): [tex]\( 977 \, \text{N} \)[/tex]
- Mass of the Earth ([tex]\( M_e \)[/tex]): [tex]\( 6 \times 10^{24} \, \text{kg} \)[/tex]
- Radius of the Earth ([tex]\( r \)[/tex]): [tex]\( 6400 \, \text{km} \)[/tex] = [tex]\( 6400 \times 10^3 \, \text{m} \)[/tex]
- Formula:
[tex]\[ W = G \frac{M_e m}{r^2} \][/tex]
Solving for [tex]\( m \)[/tex]:
[tex]\[ m = \frac{W r^2}{G M_e} \][/tex]
- Calculation:
Plugging in the values (with [tex]\( G = 6.67430 \times 10^{-11} \)[/tex]):
[tex]\[ m = \frac{977 \times (6400 \times 10^3)^2}{6.67430 \times 10^{-11} \times 6 \times 10^{24}} \][/tex]
This calculates to:
[tex]\[ m \approx 99.93037971522608 \, \text{kg} \][/tex]
#### Problem 4: Height from the Earth's Surface for Changed Weight
- Given:
- Weight at earth's surface ([tex]\( W_1 \)[/tex]): [tex]\( 300 \, \text{N} \)[/tex]
- Weight at height [tex]\( h \)[/tex] ([tex]\( W_2 \)[/tex]): [tex]\( 200 \, \text{N} \)[/tex]
- Radius of the Earth ([tex]\( r_e \)[/tex]): [tex]\( 6380 \, \text{km} \)[/tex] = [tex]\( 6380 \times 10^3 \, \text{m} \)[/tex]
- Formula:
Combine the weight formulas:
[tex]\[ W_1 = G \frac{M_e m}{r_e^2} \][/tex]
and
[tex]\[ W_2 = G \frac{M_e m}{(r_e + h)^2} \][/tex]
We get:
[tex]\[ \frac{W_2}{W_1} = \frac{r_e^2}{(r_e + h)^2} \][/tex]
Solving for [tex]\( h \)[/tex]:
[tex]\[ \frac{200}{300} = \frac{r_e^2}{(r_e + h)^2} \][/tex]
[tex]\[ \frac{2}{3} = \frac{(6380 \times 10^3)^2}{(6380 \times 10^3 + h)^2} \][/tex]
- Calculation:
Simplifying the equation, we get:
[tex]\[ (r_e + h)^2 = \frac{3}{2} r_e^2 \][/tex]
Solving for [tex]\( h \)[/tex]:
[tex]\[ h = \sqrt{\frac{3}{2} r_e^2} - r_e \][/tex]
This calculates to:
[tex]\[ h \approx 521216582732.1647 \, \text{m} \][/tex]
