Question

a newly discovered planet, \cosmo,\ has a mass that is 4 times the mass of the earth. the radius of the earth is ( r_e ). the gravitational field strength at the surface of cosmo is equal to that at the surface of the earth if the radius of cosmo is equal to
(a) ( \frac{1}{2}r_e )
(b) ( r_e )
(c) ( 2r_e )
(d) ( sqrt{r_e} )
(e) ( r_e^2 )
( g\frac{4mm}{r_e^2} )

Explanation:

Step1: Recall the formula for gravitational field strength

The gravitational field strength \(g=\frac{GM}{R^{2}}\), where \(G\) is the gravitational constant, \(M\) is the mass of the planet, and \(R\) is the radius of the planet.
For Earth, \(g_{e}=\frac{GM_{e}}{R_{e}^{2}}\). For planet Cosmo, \(g_{c}=\frac{GM_{c}}{R_{c}^{2}}\), and \(M_{c} = 4M_{e}\), \(g_{c}=g_{e}\).

Step2: Set up the equality

Since \(g_{c}=g_{e}\), we have \(\frac{GM_{e}}{R_{e}^{2}}=\frac{G\times4M_{e}}{R_{c}^{2}}\).
Cancel out \(G\) and \(M_{e}\) from both sides of the equation. We get \(\frac{1}{R_{e}^{2}}=\frac{4}{R_{c}^{2}}\).

Step3: Solve for \(R_{c}\)

Cross - multiply: \(R_{c}^{2}=4R_{e}^{2}\).
Take the square root of both sides: \(R_{c} = 2R_{e}\).

Answer:

\(C\)