Mechanics Kepler orbits

Hi,
I assume you're ok with part a) since it's just a definition but Kepler's third law is
$T^2 =kr^3$ where k is a constant.
For part b) you need to use this equation to find r for Jupiter- you have it's period, you just need k, which you can find using the data you know about Earth, because Jupiter and Earth orbit the same star.

Once you have period and radius of orbit you can find speed using the circumference of the orbit- the complicated bit here is converting the units- I assume you have the AU conversion on a datasheet.

For a circular orbit
$L=mvr$ Previous parts have given you all these figures so just plug them in.
See how you get on with those first 4 parts then tell me which of the others you're stuck on.

Hi thanks for this, I've been trying to do this problem for the last few days anyway can you explain what l is in the second equation that you stated, thank you again for the method to do part b and if you could go over the remaining parts

For part e) calculate the moment of inertia (I) using the given equation.
$L = I\omega[br]Where \omega =$angular velocity and
$\omega=\frac{2\pi}{T}$ and you have the period, so you can find angular momentum.
For part f) you're looking for Jupiter's acceleration in terms of T and r. Start with $a=\frac{v^2}{r}$ and substitute $v=\frac{2\pi r}{T}$
Use r and T to find acceleration.
For g) use the expression you found and substitute Kepler's third law into it to get an equation for centripetal acceleration only in terms of r and constants. In an orbit the centripetal acceleration is provided by gravity, so gravitational field strength is equal to centripetal acceleration. Because r is to the power of negative 2 gravitational field strength follows the inverse square law.
If you need clarification for any parts let me know

The expression you should have got in part f) is
$a=\frac{4\pi^2 r}{T^2}$
Kepler's third law states that
$T^2 =kr^3$
T^2 is in both expressions, so by substituting you get
$a=\frac{4\pi^2 r}{kr^3}$
You've got r's on the top and the bottom so it simplifies to
$a=\frac{4\pi^2 }{kr^2}$
where k is a constant for the solar system. The only part that isn't a constant is
$\frac{1}{r^2}$ so centripetal acceleration (which is the same as acceleration due to gravity in this context) follow the inverse square law with distance.