the force of gravitational attraction between the Earth and the Muskmobile.
Fgrav with Earth

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the force of gravitational attraction between the Earth and the Muskmobile. Fgrav with Earth

UpStudy AI · Accepted Answer

1. The gravitational force between two masses is given by Newton’s law of universal gravitation:  
   $$
   F_{\text{grav}} = G \frac{M m}{r^2}
   $$  
   where  
   - $$ G $$ is the gravitational constant, $$ G \approx 6.67 \times 10^{-11}\,\text{N\,m}^2/\text{kg}^2 $$,  
   - $$ M $$ is the mass of the Earth, $$ M \approx 5.97 \times 10^{24}\,\text{kg} $$,  
   - $$ m $$ is the mass of the Muskmobile, and  
   - $$ r $$ is the distance between the centers of the two masses (for a surface object, $$ r \approx $$ Earth’s radius, $$ R_E \approx 6.37 \times 10^6\,\text{m} $$).

2. Substituting the known values for an object on Earth’s surface, the equation becomes:  
   $$
   F_{\text{grav}} = G \frac{M_E m}{R_E^2}
   $$

3. Notice that the acceleration due to gravity, $$ g $$, is given by:  
   $$
   g = \frac{G M_E}{R_E^2}
   $$  
   When you compute the value:
   $$
   g \approx \frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24}}{(6.37 \times 10^6)^2} \approx 9.8\,\text{m/s}^2
   $$

4. Thus, the gravitational force on the Muskmobile is:  
   $$
   F_{\text{grav}} = m \times g \approx m \times 9.8\,\text{N}
   $$

5. If the mass of the Muskmobile is known (for example, suppose $$ m = 2000\,\text{kg} $$), then:  
   $$
   F_{\text{grav}} \approx 2000\,\text{kg} \times 9.8\,\text{m/s}^2 = 19600\,\text{N}
   $$

The final expression for the gravitational force is:  
$$
F_{\text{grav}} = G \frac{M_E m}{R_E^2} \quad \text{or} \quad F_{\text{grav}} \approx m \times 9.8\,\text{N}
$$
The gravitational force on the Muskmobile is approximately $$ 9.8 \times m $$ Newtons, where $$ m $$ is the mass of the Muskmobile.

the force of gravitational attraction between the Earth and the Muskmobile.
Fgrav with Earth

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