To calculate the gravitational force exerted by the Earth on Mahendra, we use Newton's law of universal gravitation. The formula for the gravitational force [tex]\( F \)[/tex] between two masses [tex]\( m_1 \)[/tex] and [tex]\( m_2 \)[/tex] separated by a distance [tex]\( r \)[/tex] is given by:
[tex]\[ F = G \frac{m_1 m_2}{r^2} \][/tex]
Here, the variables represent:
- [tex]\( G \)[/tex] is the gravitational constant, [tex]\( 6.67 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2 \)[/tex].
- [tex]\( m_1 \)[/tex] is the mass of the Earth, [tex]\( 6 \times 10^{24} \, \text{kg} \)[/tex].
- [tex]\( m_2 \)[/tex] is Mahendra's mass, [tex]\( 75 \, \text{kg} \)[/tex].
- [tex]\( r \)[/tex] is the radius of the Earth, [tex]\( 6.4 \times 10^6 \, \text{m} \)[/tex].
Now, let's substitute the given values into the formula:
[tex]\[ F = 6.67 \times 10^{-11} \times \frac{(6 \times 10^{24}) \times 75}{(6.4 \times 10^6)^2} \][/tex]
First, we calculate the square of the radius of the Earth:
[tex]\[ (6.4 \times 10^6)^2 = 6.4^2 \times (10^6)^2 = 40.96 \times 10^{12} = 4.096 \times 10^{13} \, \text{m}^2 \][/tex]
Next, we multiply the masses:
[tex]\[ 6 \times 10^{24} \times 75 = 450 \times 10^{24} \, \text{kg} \][/tex]
Now, substituting these values back into the formula:
[tex]\[ F = 6.67 \times 10^{-11} \times \frac{450 \times 10^{24}}{4.096 \times 10^{13}} \][/tex]
Simplify the fraction:
[tex]\[ \frac{450 \times 10^{24}}{4.096 \times 10^{13}} \approx 1.0984375 \times 10^{11} \][/tex]
So the equation becomes:
[tex]\[ F = 6.67 \times 10^{-11} \times 1.0984375 \times 10^{11} \][/tex]
Finally, multiply the constants:
[tex]\[ 6.67 \times 1.0984375 \approx 7.327880859375 \][/tex]
Therefore, we get:
[tex]\[ F \approx 7.327880859375 \times 10^{0} \][/tex]
The force is:
[tex]\[ F \approx 732.7880859375 \, \text{N} \][/tex]
Thus, the gravitational force exerted by the Earth on Mahendra is approximately [tex]\( 732.79 \, \text{N} \)[/tex].
