Certainly! Let's address this step-by-step.
### Understanding the Problem:
We know the following:
1. The mass of the heavenly body is half of the mass of the Earth.
2. The radius of the heavenly body is also half of the radius of the Earth.
3. The weight of an object on Earth is 100N (Newtons).
### Gravitational Force:
To understand how weight changes, we must look at the formula for gravitational force:
[tex]\[ F = G \frac{{m_1 \cdot m_2}}{{r^2}} \][/tex]
where:
- [tex]\( F \)[/tex] is the gravitational force (weight of the object).
- [tex]\( G \)[/tex] is the gravitational constant.
- [tex]\( m_1 \)[/tex] is the mass of the object.
- [tex]\( m_2 \)[/tex] is the mass of the heavenly body.
- [tex]\( r \)[/tex] is the radius of the heavenly body (distance between the centers of the object and the planet).
### Ratios and Changes:
Given the ratios:
1. The mass of the heavenly body, [tex]\( m_2 \)[/tex], is half the mass of the Earth ([tex]\( m_{earth} \)[/tex]).
2. The radius, [tex]\( r \)[/tex], of the heavenly body is half the radius of the Earth ([tex]\( r_{earth} \)[/tex]).
### Finding the New Weight:
1. Mass Ratio: The mass of the heavenly body is [tex]\( \frac{1}{2} \)[/tex] the Earth's mass.
2. Radius Ratio: The radius of the heavenly body is [tex]\( \frac{1}{2} \)[/tex] the Earth's radius.
### Gravitational Force Proportionality:
Gravitational force is proportional to the mass divided by the square of the radius:
[tex]\[ F \propto \frac{m}{r^2} \][/tex]
So, if we compare the gravitational force [tex]\( F_h \)[/tex] on the heavenly body to the gravitational force [tex]\( F_e \)[/tex] on Earth:
[tex]\[ \frac{F_h}{F_e} = \frac{\left( \frac{1}{2} m_{earth} \right)}{\left( \frac{1}{2} r_{earth} \right)^2} \][/tex]
### Simplifying the Ratio:
- [tex]\( m_{earth} \)[/tex] cancels out in the numerator.
- [tex]\( r_{earth} \)[/tex] cancels out in the denominator.
Thus:
[tex]\[ \frac{F_h}{F_e} = \frac{ \left( \frac{1}{2} \right) }{ \left( \frac{1}{2} \right)^2 } \][/tex]
[tex]\[ \frac{F_h}{F_e} = \frac{ \frac{1}{2} }{ \frac{1}{4} } \][/tex]
[tex]\[ \frac{F_h}{F_e} = \frac{1}{2} \times \frac{4}{1} \][/tex]
[tex]\[ \frac{F_h}{F_e} = 2 \][/tex]
### Conclusion:
Since the weight on Earth [tex]\( F_e \)[/tex] is 100N:
[tex]\[ F_h = 2 \times F_e \][/tex]
[tex]\[ F_h = 2 \times 100N \][/tex]
[tex]\[ F_h = 200N \][/tex]
Thus, the weight of the object on the heavenly body would be 200N.
