Suppose all the mass of the Earth was compacted into a small spherical ball. What radius must the sphere have so that the acceleration due to gravity at the Earth’s new surface would equal the acceleration due to gravity at the surface of the Sun?

It is given that the mass of Earth compacted into a small spherical ball and setting

the acceleration due to the gravity of Earth is equal to the acceleration due to the gravity of the Sun. Therefore,

\(\begin{aligned}{g_E} &= {g_S}\\\frac{{G{M_E}}}{{{R^2}}} &= \frac{{G{M_S}}}{{R_s^2}}\\\frac{{{M_E}}}{{{R^2}}} &= \frac{{{M_S}}}{{R_s^2}}\end{aligned}\)

Where \({M_E}\) is the mass of Earth (\(5.98 \times {10^{24}}\;{\rm{kg}}\)), \({M_S}\) is the mass of the Sun (\(2 \times {10^{30}}\;{\rm{kg}}\)), \({R_S}\) is the solar radius (\(6.955\; \times {10^8}\;{\rm{m}}\))

\(R = \)the radius of the new sphere

\(\begin{aligned}{R^2} &= \frac{{{M_E}R_S^2}}{{{M_S}}}\\ &= \frac{{\left( {5.98 \times {{10}^{24}}\;{\rm{kg}}} \right) \times {{\left( {6.955 \times {{10}^8}\;{\rm{m}}} \right)}^2}}}{{2 \times {{10}^{30}}\;{\rm{kg}}}}\\ &= \;1.446\; \times \;{10^{12}}\;{{\rm{m}}^2}\end{aligned}\)

Therefore, the radius is given by:

\(\begin{aligned}R &= \;\sqrt {1.446\; \times \;{{10}^{12}}\;{{\rm{m}}^{\rm{2}}}} \\ &= \;1.20 \times \;{10^6}\;{\rm{m}}\end{aligned}\)

So, the radius of the new sphere is \(R = 1.20 \times {10^6}\;{\rm{m}}\).