Question

Which of the following changes to the earth-sun system would reduce the magnitude of the force between them to one-fourth the value found in Part A?
Reduce the mass of the earth to one-fourth its normal value.
Reduce the mass of the sun to one-fourth its normal value.
Reduce the mass of the earth to one-half its normal value and the mass of the sun to one-half its normal value.
Increase the separation between the earth and the sun to four times its normal value.

Answer 1

Answer:

-Reduce the mass of the earth to one-fourth its normal value.

-Reduce the mass of the sun to one-fourth its normal value.

-Reduce the mass of the earth to one-half its normal value and the mass of the sun to one-half its normal value.

Explanation:

The force (F) between two massive bodies (earth-sun system) is given by the following equation:

$F=\frac{g*M_{e}*M_{s}}{r^{2}}$

Where "g" is the gravitational constant "Me" is Earth's mass, "Ms" is the Sun's mass and "r" is the separation between the Earth and the Sun.

1-) If we reduce the mass of the earth to one-fourth its normal value:

$F^{*} =\frac{g*\frac{M_{e}}{4} *M_{s}}{r^{2}}\\F^{*} =\frac{g*{M_{e}} *M_{s}}{4r^{2}}\\F^{*} =\frac{F}{4}$

2-) Reduce the mass of the sun to one-fourth its normal value.

$F^{*} =\frac{g*\frac{M_{s}}{4} *M_{e}}{r^{2}}\\F^{*} =\frac{g*{M_{s}} *M_{e}}{4r^{2}}\\F^{*} =\frac{F}{4}$

3-) Reduce the mass of the earth to one-half its normal value and the mass of the sun to one-half its normal value.

$F^{*} =\frac{g*\frac{M_{s}}{2}*\frac{M_{e} }{2}}{r^{2}}\\F^{*} =\frac{g*{M_{s}} *M_{e}}{(2*2)r^{2}}\\F^{*} =\frac{F}{4}$

4-) Increase the separation between the earth and the sun to four times its normal value.

$F^{*}=\frac{g*M_{e}*M_{s}}{(4r)^{2}}\\F^{*}=\frac{g*M_{e}*M_{s}}{16r^{2}}\\F^{*}=\frac{F }{16}$.

Therefore, changes 1, 2 and 3 would reduce the magnitude of the force between them to one-fourth of the original value.