$$
begin{align*}
v&=sqrt{frac{m_ecdot{G}}{r}}
end{align*}
$$

Where the $m_e$ stands for the center of the Earth as the satellite orbits around the Earth, $G$ is universal gas constant $G=6.67cdot{10^{-11}} frac{text{N}cdottext{m}^2}{text{kg}^2}$. And $r$ is distance between the center of the Earth and the center of satelite.

If the satellite height increase for $24text{ km}$, the new distance $r$ will be:

$$
begin{align*}
r&=r_e+h\
r&=6.38cdot{10^6}text{ m}+225cdot{10^3}text{ m}+24cdot{10^3}text{ m}\
r&=6629cdot{10^3}text{ m}
end{align*}
$$

Let’s substitute and compute the velocity:

$$
begin{align*}
v&=sqrt{frac{m_ecdot{G}}{r}}\
v&=sqrt{frac{5.97cdot{10^{24}text{ kg}}cdot{6.67cdot{10^{-11}} frac{text{N}cdottext{m}^2}{text{kg}^2}}}{6629cdot{10^3}text{ m}}}\
v&=7750 frac{text{m}}{text{s}}
end{align*}
$$

$$
boxed{text{The satellite is slower.}}
$$

$$
begin{align*}
v&=sqrt{frac{m_ecdot{G}}{r}}
end{align*}
$$

Where the $m_e$ stands for the center of the Earth as the satellite orbits around the Earth, $G$ is universal gas constant $G=6.67cdot{10^{-11}} frac{text{N}cdottext{m}^2}{text{kg}^2}$. And $r$ is distance between the center of the Earth and the center of satelite.

If the satellite height is $150text{ km}$, the distance between the Earth and satellite center is

$$
begin{align*}
r&=r_e+h\
r&=6.38cdot{10^6}text{ m}+150cdot{10^3}text{ m}\
r&=6530cdot{10^3}text{ m}
end{align*}
$$

Let’s substitute and compute the velocity:

$$
begin{align*}
v&=sqrt{frac{m_ecdot{G}}{r}}\
v&=sqrt{frac{5.97cdot{10^{24}text{ kg}}cdot{6.67cdot{10^{-11}} frac{text{N}cdottext{m}^2}{text{kg}^2}}}{6530cdot{10^3}text{ m}}}
end{align*}
$$

$$
boxed{v=7800 frac{text{m}}{text{s}}}
$$

$bold{b)}$

The period needed to pass one orbit is given by the equation:

$$
begin{align*}
T&=2cdotpisqrt{frac{r^3}{m_ecdot{G}}}
end{align*}
$$

We have all data, so let’s substitute and compute:

$$
begin{align*}
T&=2cdotpisqrt{frac{(6530cdot{10^3}text{ m})^3}{5.97cdot{10^{24}text{ kg}}cdot{6.67cdot{10^{-11}} frac{text{N}cdottext{m}^2}{text{kg}^2}}}}\
T&=2cdotpicdot{836.2text{ s}}\
T&=5254text{ s}
end{align*}
$$

$$
boxed{T=87text{ min}text{ }34text{ s}}
$$

b) $T=87text{ min}text{ }34text{ s}$

$$
begin{align*}
v&=sqrt{frac{m_mcdot{G}}{r}}
end{align*}
$$

Where the $m_m$ stands for the mass of the Mercury as the satellite orbits around the Mercury, $G$ is universal gas constant $G=6.67cdot{10^{-11}} frac{text{N}cdottext{m}^2}{text{kg}^2}$. And $r$ is distance between the center of the Mercury and the center of satelite.

Let’s find the data for Mercury from the table $17-1$:

$$
begin{align*}
r_m&=2.44cdot{10^6}text{ m}\
m_m&=3.3cdot{10^{23}}text{ kg}
end{align*}
$$

If the satellite height is $260text{ km}$, the distance between the Mercury and satellite center is

$$
begin{align*}
r&=r_m+h\
r&=2.44cdot{10^6}text{ m}+260cdot{10^3}text{ m}\
r&=2700cdot{10^3}text{ m}
end{align*}
$$

Let’s substitute and compute the velocity:

$$
begin{align*}
v&=sqrt{frac{m_mcdot{G}}{r}}\
v&=sqrt{frac{3.3cdot{10^{23}}text{ kg}cdot{6.67cdot{10^{-11}} frac{text{N}cdottext{m}^2}{text{kg}^2}}}{2700cdot{10^3}text{ m}}}
end{align*}
$$

$$
boxed{v=2855 frac{text{m}}{text{s}}}
$$

The period needed to pass one orbit is given by the equation:

$$
begin{align*}
T&=2cdotpisqrt{frac{r^3}{m_mcdot{G}}}
end{align*}
$$

We have all data, so let’s substitute and compute:

$$
begin{align*}
T&=2cdotpisqrt{frac{(2700cdot{10^3}text{ m})^3}{3.3cdot{10^{23}text{ kg}}cdot{6.67cdot{10^{-11}} frac{text{N}cdottext{m}^2}{text{kg}^2}}}}\
T&=2cdotpicdot{945text{ s}}\
T&=5937text{ s}
end{align*}
$$

$$
boxed{Tapprox{99text{ min}}}
$$

b) $Tapprox{99text{ min}}$