What is the surface gravity on this distant planet?

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Homework Help Overview

The discussion revolves around calculating the surface gravity on a distant planet, given its mass and radius in relation to Earth. The original poster presents measurements indicating the planet's mass is twice that of Earth, but the free-fall acceleration is one-fourth that of Earth's. The radius is specified as 1.8 x 107 meters.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of gravitational equations and check the correctness of the provided radius and mass. There are attempts to calculate escape velocity and surface gravity, with some questioning the accuracy of numerical values and the setup of equations.

Discussion Status

The discussion is ongoing, with participants providing feedback on calculations and sharing their own experiences of similar mistakes. There is a focus on verifying the values used in equations and ensuring the correct application of gravitational principles.

Contextual Notes

Participants note the importance of correctly interpreting the given values and equations, with some expressing uncertainty about their calculations. The conversation reflects a collaborative effort to clarify the problem without reaching a definitive conclusion.

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Homework Statement


You have been visiting a distant planet. Your measurements have determined that the planet's mass is twice that of Earth but the free-fall acceleration at the surface is only one-fourth as large. Radius = 1.8*10^7meters
Mass = 1.19484 × 10^25 kilograms


Homework Equations


U_g = -(G*M*m)/r
v_escape = sqrt(2GM/r)

The Attempt at a Solution



I just plugged in the variables to the espace velocity equation and get 94km/s, but that's incorrect.
 
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The radius was given in the first part of the problem.
 
Um. The radius and the mass are correct. But I get 9.4km/sec. Check the decimal point.
 
Feldoh said:
g_e = \frac{GM_{e}}{R_{e}}

\frac{1}{4}g_e = g_{distant planet}

\frac{1}{4}*\frac{GM_{e}}{R_{e}} = \frac{G*2M_{e}}{R_{distant planet}}

\frac{1}{4*R_{e}} = \frac{2}{R_{distant planet}}

R_{distant planet} = 8R_{e}

g=G*M/r^2. It's force/mass not potential/mass.
 
Yeah, looks like I was just being retarded
 
Feldoh said:
Yeah, looks like I was just being retarded

S'ok. Reminds me of my post where I was strongly suggesting an OP integrate voltage over a gaussian surface to get enclosed charge. Duh.
 
Dick said:
S'ok. Reminds me of my post where I was strongly suggesting an OP integrate voltage over a gaussian surface to get enclosed charge. Duh.

Lol, guess we all screw up sometimes :smile:
 

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