Is the type of a knot dependent on the space in which it is embedded or not? For example in this image a satellite construction is depicted. On the left you have a knot embedded in a solid torus. Is this the trivial knot or not? If this same exact diagram was drawn for a knot in $\mathbb{R}^3$ the knot would be unknotted. The problem is that the embedding itself dictates the equivalence class of the knot. So in this case what is the equivalence class of the knot in the torus? 
 A: One thing to clear up: what is a knot?  One definition is a circle embedded in $S^3$ (or $\mathbb{R}^3$) considered up to isotopy, in which case the circle embedded in a solid torus is not a knot.  Another definition is a circle embedded in a manifold considered up to isotopy, in which case the knot in the solid torus is (by definition) inequivalent to a knot embedded in any other space.
Let's say that a knot is trivial if it bounds a disk in the complement.  You ask whether the knot in the solid torus is trivial.
Theorem. There is a one-to-one correspondence between links of the following two kinds.  (1) A link in a solid torus with marked longitude on the boundary. (2) A link in $S^3$, one of whose components is a distinguished unknot.
Proof. Start with a link as described in (2).  Isotope the link so that the distinguished unknot is the $z$-axis, passing through $\infty$.  Remove a tubular neighborhood of this unknot, and the rest of the link is a link in a solid torus, with a meridian of the unknot being the marked longitude.  Conversely, a link as in (1) can become a link as in (2) by filling in the complement with a solid torus to get a sphere (with the meridian of the additional solid torus being glued to the marked longitude), the core of which is the additional unknot.  Link equivalence in one representation carries over as link equivalence in the other (proof left as exercise). Q.E.D.
If a knot in a solid torus is trivial then its $S^3$ representation is a split unlink.  The marked longitude is meant to deal with Dehn twists of the solid torus.  If a knot in a solid torus is trivial, it is trivial no matter the choice of marked longitude
If we add the unlink to your knot in the solid torus, we get a Whitehead link.  This link is not a split unlink, for instance by looking at the Jones polynomial.
