Change the chapter
Question
Earth is $1.5\times 10^{11}\textrm{ m}$ from the Sun. Mercury is $5.7\times 10^{ 10}\textrm{ m}$ from the Sun. How does the gravitational field of the Sun on Mercury ($g_{SM}$) compare to the gravitational field of the Sun on Earth ($g_{SE}$)?
Question by OpenStax is licensed under CC BY 4.0.
Final Answer
The graviational field of the sun at Mercury is greater than that at Earth by a factor of 6.9.
Solution Video

OpenStax College Physics for AP® Courses Solution, Chapter 6, Problem 6 (Test Prep for AP® Courses) (1:14)

Sign up to view this solution video!

View sample solution

Calculator Screenshots

OpenStax College Physics, Chapter 6, Problem 6 (AP) calculator screenshot 1
Video Transcript

This is College Physics Answers with Shaun Dychko. The gravitational field due to the Sun is gonna be measured by the gravitational constant times the mass of the Sun divided by the distance from the Sun squared. So in this case, we have the distance to Mercury and in the case of the acceleration due to gravity of the Sun on Earth, it's the same numerator but we are dividing by the distance to the Earth squared. And so to find... in order to compare these two, let's divide them. So we have <i>g SM</i> divided by <i>g SE</i> and so we are taking this fraction and copying it here and then instead of dividing a fraction by a fraction, I am instead multiplying by the reciprocal of <i>G SE</i> this is being in the denominator here but I don't wanna put this fraction in the denominator so I'm instead multiplying by its reciprocal. And these <i>G</i>'s will cancel and mass of the Sun will cancel and we are left with distance to the Earth divided by distance to Mercury squared. So it's 1.5 times 10 to the 11 meters— distance to Earth— divided by 5.7 times 10 to the 10 meters— distance from the Sun to Mercury— and squared gives 6.9. So the gravitational field of the Sun at Mercury is greater than that at Earth by a factor of 6.9.