Question

if the acceleration due to gravity on the earth is g, which formula gives the acceleration due to gravity on loput? view available hint(s) o ( g\frac{1.7}{5.6} ) o ( g\frac{1.7^{2}}{5.6} ) o ( g\frac{1.7^{3}}{5.6^{2}} ) o ( g\frac{5.6}{1.7} ) o ( g\frac{5.6^{2}}{1.7^{2}} ) o ( g\frac{5.6}{1.7^{2}} ) submit

Explanation:

1. Explanation:

• The acceleration due to gravity \(g\) is given by the formula \(g=\frac{GM}{R^{2}}\), where \(G\) is the gravitational - constant, \(M\) is the mass of the planet, and \(R\) is the radius of the planet. Let the mass of the Earth be \(M_{E}\), radius be \(R_{E}\), and the acceleration due to gravity be \(g_{E}=g = \frac{GM_{E}}{R_{E}^{2}}\). Suppose the mass of the planet Loput is \(M_{L}\) and radius is \(R_{L}\), and the acceleration due to gravity on Loput is \(g_{L}=\frac{GM_{L}}{R_{L}^{2}}\). If we assume \(M_{L}=1.7M_{E}\) and \(R_{L}=5.6R_{E}\), then \(g_{L}=\frac{G(1.7M_{E})}{(5.6R_{E})^{2}}\).
• Since \(g = \frac{GM_{E}}{R_{E}^{2}}\), we can substitute \(GM_{E}=gR_{E}^{2}\) into the formula for \(g_{L}\). So \(g_{L}=\frac{1.7gR_{E}^{2}}{5.6^{2}R_{E}^{2}}\).
• The \(R_{E}^{2}\) terms cancel out, and we get \(g_{L}=g\frac{1.7}{5.6^{2}}\).

1. Answer: \(g\frac{1.7}{5.6^{2}}\) (corresponding to the third option in the multiple - choice list)

Answer:

1. Explanation:

• The acceleration due to gravity \(g\) is given by the formula \(g=\frac{GM}{R^{2}}\), where \(G\) is the gravitational - constant, \(M\) is the mass of the planet, and \(R\) is the radius of the planet. Let the mass of the Earth be \(M_{E}\), radius be \(R_{E}\), and the acceleration due to gravity be \(g_{E}=g = \frac{GM_{E}}{R_{E}^{2}}\). Suppose the mass of the planet Loput is \(M_{L}\) and radius is \(R_{L}\), and the acceleration due to gravity on Loput is \(g_{L}=\frac{GM_{L}}{R_{L}^{2}}\). If we assume \(M_{L}=1.7M_{E}\) and \(R_{L}=5.6R_{E}\), then \(g_{L}=\frac{G(1.7M_{E})}{(5.6R_{E})^{2}}\).
• Since \(g = \frac{GM_{E}}{R_{E}^{2}}\), we can substitute \(GM_{E}=gR_{E}^{2}\) into the formula for \(g_{L}\). So \(g_{L}=\frac{1.7gR_{E}^{2}}{5.6^{2}R_{E}^{2}}\).
• The \(R_{E}^{2}\) terms cancel out, and we get \(g_{L}=g\frac{1.7}{5.6^{2}}\).

1. Answer: \(g\frac{1.7}{5.6^{2}}\) (corresponding to the third option in the multiple - choice list)