What would be its weight if the earth had twice its present radius and twice its present mass?

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Concept:

Newton's law of gravitation:

• "Any particle of matter in the universe attracts any other with a force varying directly as the product of the masses and inversely as the square of the distance between them". 
• It is directed along the line joining the point masses.
• It is a conservative force field, i.e mechanical energy remains conserved.
• It is a central force i.e angular momentum remains conserved.

For two mass m1 and m2, it is given by

\(F = \frac{Gm_1 \times m_2}{r^2}\)

• Gravity Equation or Newton’s universal law of gravitation:

• where G is the gravitational constant, F is the force due to gravity between two masses (m1 and m2), which are a distance r apart.
• The expression for acceleration due to gravity (g) on the surface of a planet is given by:

F = mg

\(mg= {GMm \over r^2}\)

\(g= {GM \over r^2}\)

where r is the radius of the planet; G is the gravitational constant, M is the mass of the planet.

Calculation:

Given that,

Final mass of earth, M’ = 2 × initial mass of earth, 2M

Final radius of earth, R’ = 2× initial radius of earth, 2R

Thus, the ratio of final acceleration and initial acceleration can be given as

\(\frac{{g'}}{g} = \frac{{\frac{{GM'}}{{{R^{'2}}}}}}{{\frac{{GM}}{{{R^2}}}}} = \frac{{\frac{{2GM}}{{ {{\left( {2R} \right)}^2}}}}}{{\frac{{GM}}{{{R^2}}}}} = \frac{1}{2}\)

\(\therefore g' = \frac{g}{2}\)

Hence the weight becomes half of the original weight.

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