There is no hope. Timothy Budd already explained why staying on the sphere $S_t$ of possible locations of $M$ will not work; so this is to explain why leaving it will not help. What is enough to show is the following claim: at large times, if $t_1<t_2$ and $C_{t_{1,2}}$ are at $\epsilon$-neighborhoods of $S_{t_{1,2}}$ respectively, then the radial projections of $C_{t_{1,2}}$ onto the unit sphere are at distance at most $c\cdot \log {t_2/t_1}$ from each other. With the claim, essentially the same argument will work.

But the claim is fairly obvious: between $t_1$ and $t_2$, $C_t$ cannot get more than $(v_c-v_m)(t_2-t_1)+2\epsilon$ behind $S_t$, otherwise it wouldn't be able to catch up. If $t_2-t_1<t_1\cdot \frac{v_c-v_m}{100v_m}$, then this means that $C_t$ is always at distance at least $\frac12v_m t$ from the origin, and then the speed of the projection is bounded by $\mathrm{const}/t$ as in Budd's answer. If $t_2/t_1\geq 1+\frac{v_c-v_m}{100v_m}$, then we can just bound the distance between projections of $C_{t_{1,2}}$ by $2$, and take the constant $c$ to be $2\cdot \log\left(1+\frac{v_c-v_m}{100v_m}\right)$

There is no hope. Timothy Budd already explained why staying on the sphere $S_t$ of possible locations of $M$ will not work; so this is to explain why leaving it will not help. What is enough to show is the following claim: at large times, if $t_1<t_2$ and $C_{t_{1,2}}$ are at $\epsilon$-neighborhoods of $S_{t_{1,2}}$ respectively, then the radial projections of $C_{t_{1,2}}$ onto the unit sphere are at distance at most $c\cdot \log {t_2/t_1}$ from each other. With the claim, essentially the same argument will work.

(Details: think about the task in terms of gradually ruling out possible directions in which $M$ can go. One can only exclude a direction if at some time $t$, $C$ is at the $\epsilon$-neighborhood of the corresponding point of $S_t$. Suppose Alice has a radar that she can direct to any point of $S_t$ and rule out the $2\epsilon$-neighborhood of it. Let Bob follow any strategy, and let Alice do this: whenever Bob's $C$ is in the closed $\epsilon$-neighborhood of $S_t$, she directs the radar to the projection of Bob's $C$ on $S_t$; in between those times, she turns her radar at constant speed. Clearly Alice will know more than Bob, yet by the claim, the angular speed of the radar will be bounded by $c/t$, and by Budd's argument, Alice will not be able to rule out everything.)

But the claim is fairly obvious: between $t_1$ and $t_2$, $C_t$ cannot get more than $(v_c-v_m)(t_2-t_1)+2\epsilon$ behind $S_t$, otherwise it wouldn't be able to catch up. If $t_2-t_1<t_1\cdot \frac{v_c-v_m}{100v_m}$, then this means that $C_t$ is always at distance at least $\frac12v_m t$ from the origin, and then the speed of the projection is bounded by $\mathrm{const}/t$ as in Budd's answer. If $t_2/t_1\geq 1+\frac{v_c-v_m}{100v_m}$, then we can just bound the distance between projections of $C_{t_{1,2}}$ by $2$, and take the constant $c$ to be $2\cdot \log\left(1+\frac{v_c-v_m}{100v_m}\right)$

There is not hope. Timothy Budd already explained why staying on the sphere $S_t$ of possible locations of $M$ will not work; so this is to explain why leaving it will not help. What is enough to show is the following claim: at for large times, if $t_1<t_2$ and $C_{t_{1,2}}$ is at $\epsilon$-neighborhood of $S_{t_{1,2}}$, then the radial projections of $C_{t_{1,2}}$ onto the unit sphere are at distance at most $c\cdot \log {t_2/t_1}$ from each other. With the claim, essentially the same argument will work.

But the claim is fairly obvious: between $t_1$ and $t_2$, $C_t$ cannot get more than $(v_c-v_m)(t_2-t_1)+2\epsilon$ behind $S_t$, otherwise it wouldn't be able to catch up. If $t_2-t_1<t_1\cdot \frac{v_c-v_m}{100v_m}$, then this means that $C_t$ is always at distance at least $\frac12v_m t$ from the origin, and then the speed of the projection is bounded by $\mathrm{const}/t$ as in Budd's answer. If $t_2/t_1\geq 1+\frac{v_c-v_m}{100v_m}$, then we can just bound the distance between projections of $C_{t_{1,2}}$ by $2$, and take the constant $c$ to be $2\cdot \log\left(1+\frac{v_c-v_m}{100v_m}\right)$

There is no hope. Timothy Budd already explained why staying on the sphere $S_t$ of possible locations of $M$ will not work; so this is to explain why leaving it will not help. What is enough to show is the following claim: at large times, if $t_1<t_2$ and $C_{t_{1,2}}$ are at $\epsilon$-neighborhoods of $S_{t_{1,2}}$ respectively, then the radial projections of $C_{t_{1,2}}$ onto the unit sphere are at distance at most $c\cdot \log {t_2/t_1}$ from each other. With the claim, essentially the same argument will work.

But the claim is fairly obvious: between $t_1$ and $t_2$, $C_t$ cannot get more than $(v_c-v_m)(t_2-t_1)+2\epsilon$ behind $S_t$, otherwise it wouldn't be able to catch up. If $t_2-t_1<t_1\cdot \frac{v_c-v_m}{100v_m}$, then this means that $C_t$ is always at distance at least $\frac12v_m t$ from the origin, and then the speed of the projection is bounded by $\mathrm{const}/t$ as in Budd's answer. If $t_2/t_1\geq 1+\frac{v_c-v_m}{100v_m}$, then we can just bound the distance between projections of $C_{t_{1,2}}$ by $2$, and take the constant $c$ to be $2\cdot \log\left(1+\frac{v_c-v_m}{100v_m}\right)$