You are here:

Astrophysics/Help with planet accelaration

Advertisement


Question
QUESTION: HI-I'm helping a friend with a question. I'm not asking you to solve it-I'm asking you to please show me how to set it up so I can find the solution-I am stumped-please help:

What is the acceleration of Jupiter produced by the gravitational force ?
When 'g' between jupiter and the sun is 4.19 * 10 ^23 N, and Jupiter's orbital radius is 778.3 x 10^9 m.
If possible use the same information to calculate the acceleration of the Sun.

Thank you.
Jason


ANSWER: Hello,

This problem is extremely straightforward.

However, what you're calling 'g' is, in fact, NOT g (as in an 'acceleration of gravity') but rather **mg** (the equivalent of a 'weight' for Jupiter as referenced on the Sun- since the units are Newtons, N *not* m/s^2!)

Once that is understood, it all falls into place.

The solution then quickly follows (or should) once you recognize the centripetal acceleration of Jupiter (which is what you want) is being provided by the Sun, so:

F (force of attraction) = G (M(s) m/ r^2 = mv^2/ r  = mg

where m is Jupiter's mass and M(s) is the Sun's (~ 2 x 10^30 kg)

From here it is, as we say, a "piece of cake" - or ought to be for you!



---------- FOLLOW-UP ----------

QUESTION: Hi, and thank you. So m is 4.19 * 10 ^23 N, Jupiter's mass. What is G?
The gravitational constant, G = 6.67428 * 10^(-11) N(m/kg)? I just want to be sure.
Please let me know. Thank you.

Jason G.


Answer
Hello,

NO...m is NOT 4.19 * 10 ^23 N, (Jupiter's mass). It is the putative FORCE (mg) - by which Jupiter is attracted to the Sun. (Note the units for mass are in kg for the SI system, and *force* is given in N).


G is the gravitational constant as you have defined it (sorry, I thought I'd included it!)


Now, since we know:

G (M(s) m/ r^2 = mv^2/ r  = mg

and have G, M(s) and r, we can compute either m or g. (m would be found by dividing 4.19 * 10 ^23 N by the value obtained for g.)

But since the problem specifically wants the *acceleration* (v^2/r) we can see that is exactly g.

Then: v^2/ r = g  = G (M(s)/ r^2

Which you should be able to work out from here.  

Astrophysics

All Answers


Answers by Expert:


Ask Experts

Volunteer


Philip A. Stahl

Expertise

I specialize in stellar and solar astrophysics. Can answer any questions pertaining to these areas, the spectroscopic analysis of stars – as well as the magneto-hydrodynamics of sunspots and solar flares. Sorry – No homework problems done or research projects! I will provide hints on solutions.

Experience

Have published papers on the relationship between sunspot morphology and solar flares; discovery of SID flares related to this, constructed computerized stellar models; MHD research.

Organizations
American Astronomical Society (Solar physics and Dynamical astronomy divisions), American Geophysical Union, American Mathematical Society, Intertel.

Publications
Solar Physics, Journal of the Royal Astronomical Society of Canada, Journal of the Barbados Astronomical Society, Meudon Solar Flare Proceedings (Meudon, France). Books: 'Selected Analyses in Solar Flare Plasma Dynamics', 'Physics Notes for Advanced Level'.

Education/Credentials
B.A. degree in Astronomy; M.Phil. degree in Physics - specializing in solar physics.

Awards and Honors
Postgraduate research award- Barbados government; Studentship Award in Solar Physics - American Astronomical Society

©2012 About.com, a part of The New York Times Company. All rights reserved.