Calculating Force and Acceleration in Circular Motion Problem

[phys1618]

Homework Statement

If John was stationary in space at eh same distance fromo the sun as the Earth is, a.how much force would the sun be exeerting on John? b.what would be John's acceleration towards the sun? c.what would be the sun's acceleration toward John??

Homework Equations

Mass of John= 50kg
Mass of Sun = 2x10³⁰kg
r=1.5x10¹¹m
T=331536000s
Fg=G*M_sun M_john/r²
a=4pi²r/T²
F_s=force of the sun
F_j=force of john
a_s=acceleration of sun
a_j=acceleration of john

The Attempt at a Solution

a.
Fs=GMsMj/r²
Fs=(6.67X10^-11)(2x10³⁰)(50)/(1.5x10¹¹)²
Fs= .2964444444 N

b. aj=4pi²r/T²
aj=4pi²(1.5x10¹¹m)/(31536000s)²
aj= .0059497487m/s²

c. Would I use the same equation as b, with the same T? or is it a different T??

thank you for all helps and comments

 

[alphysicist]

Hi phys1618,

The Attempt at a Solution

a.
Fs=GMsMj/r²
Fs=(6.67X10^-11)(2x10³⁰)(50)/(1.5x10¹¹)²
Fs= .2964444444 N

b. aj=4pi²r/T²
aj=4pi²(1.5x10¹¹m)/(31536000s)²
aj= .0059497487m/s²

This is right, but there is a more direct method that will also help you in part c. In part a you found the force; what is the relationship between force and acceleration? You should get the same answer.

Then for part c, do the same procedure to find the acceleration for the sun (which will of course be much smaller).

 

[phys1618]

F=ma
yea, it did gave me the same answer as b
so for c. I
Fs=Msun*a of the sun
.2964444444 N=2x10^30kg* a
1.48 x 10^-31 m/s^2=a
right??

thank you for the help!

 

[alphysicist]

Sure, glad to help!