Question

A satellite is launched into an orbit around the Earth. It has a critical velocity of about $6{\text{km s}}^{-1}$. What is the height of the satellite in $\text{km}$? ($\text{G}=6.67\times {10}^{-11}{\text{Nm}}^2{\text{kg}}^{-2}$, $\text{M}=6\times {10}^{24}\text{kg}$, and $\text{R}=6,370\text{km}$)

• A. 4745.67

Answer 1

The correct option is A 4745.67
Given:
$\text{G}=6.67\times {10}^{-11}{\text{Nm}}^2{\text{kg}}^{-2}$
$\text{M}=6\times {10}^{24}\text{kg}$
$\text{R}=6,370\text{km}$
${\text{v}}_{\text{esc}}=6{\text{km s}}^{-1}$

${\text{v}}_{\text{esc}}=\sqrt{\frac{\text{GM}}{\text{R}+\text{h}}}$

$\text{R}+\text{h}=\frac{\text{GM}}{{\text{v}}_{\text{esc}}^2}$

$=\frac{6.67\times {10}^{-11}\times 6\times {10}^{24}}{(6\times {10}^3{)}^2}$

$\text{R}+\text{h}=11,116.67\text{km}$
$\text{h}=11,116.67-6,371$
$=4,745.67\text{km}$