91Ó°ÊÓ

The given data can be listed below as,

• The distance of perigee from the earth’s surface is,$h_p=400\text{ k³¾}$.
• The distance of apogee from the earth’s surface is,role="math" localid="1655735322547" $h_a=4000\text{ k³¾}$.

The total amount of energy of an isolated system remains stable, even though internal changes occur when energy disappears in one form and appears in another.

a)

The spacecraft moving in an elliptical orbit, the relation between the time period, the mass of earth, and the semi-major axis is expressed as,

$T=\frac{2\mathrm{Ï€}{a}^{3/2}}{\sqrt{G{M}_E}}$ ...(i)

Here $T$is the time period, role="math" localid="1655735444026" $a$is the semi-major axis of an elliptical orbit, $G$is the gravitational constant, and $M_E$is the mass of the sun.

The major axis of an elliptical orbit is expressed as,

$2a=r_a+r_p$ ...(ii)

Hererole="math" localid="1655735530960" $2a$is the major axis of an elliptical orbit,role="math" localid="1655735628136" $r_a$is the radius of apogee and$r_p$is the radius of perigee. The radius of apogee and the radius of perigee are expressed as,

$r_a=h_a+R_E$

And

$r_p=h_p+R_E$

Here$R_E$is the radius of the earth.

Substitute the value of $r_a$and $r_p$in equation (ii)

Substitute$6.37×{10}^6\text{″¾}$ for$R_E$ ,$\text{400}×{\text{10}}^3\text{″¾}$ for$h_p$ , and$\text{4000}×{\text{10}}^3\text{″¾}$ for$h_a$ in the above equation.

$\begin{array}{c}a=6.37×{10}^6\text{″¾}+\frac{(\text{400}×{\text{10}}^3\text{″¾}+\text{4000}×{\text{10}}^3\text{″¾})}{2}\\ =8.57×{10}^6\text{″¾}\end{array}$

Substitute$6.673×{10}^{−11}{\text{ N³¾}}^{\text{2}}/{\text{kg}}^{\text{2}}$ for $G$,$8.57×{10}^6\text{″¾}$ for$a$ , and$5.97×{10}^{24}\text{ k²µ}$ for$M_E$ in the equation (i).

$\begin{array}{c}T=\frac{2\mathrm{Ï€}×{(8.57×{10}^6\text{″¾})}^{\frac{3}{2}}}{\sqrt{6.673×{10}^{−11}{\text{ N³¾}}^{\text{2}}/{\text{kg}}^{\text{2}}×5.97×{10}^{24}\text{ k²µ}}}\\ =7.9×{10}^3\text{ s}\end{array}$

Hence the time period of spacecraft orbit is,$7.9×{10}^3\text{ s}$ .

b)

The conservation of angular momentum between apogee and perigee is expressed as,

Here$V_P$ is the speed of perigee and$V_a$ is the speed of apogee.

Substitute the value of $r_p$and$r_a$ in the above equation.

$\frac{v_p}{v_a}=\frac{h_a+R_E}{h_p+R_E}$

Substitute$6.37×{10}^6\text{″¾}$ for $R_E$,$\text{400}×{\text{10}}^3\text{″¾}$ for $h_p$, and $\text{4000}×{\text{10}}^3\text{″¾}$for$h_a$ in the above equation.

Hence the ratio of the spacecraft’s speed at perigee to its speed at apogee is, 1.53.

c)

The conservation of energy to apogee and perigee is expressed as

$K_a+U_a=K_p+U_p$ ...(iii)

Here$K_a$and $K_P$are the kinetic energy of apogee and perigee androle="math" localid="1655736678099" $U_a$and$U_P$are the gravitational potential energy of apogee and perigee.

The kinetic and potential energy of apogee is expressed as,

$K_a=\frac{1}{2}m{v}_a^2$

And

$U_a=\frac{G{M}_{E}m}{r_a}$

The kinetic and potential energy of perigee is expressed as,

$K_p=\frac{1}{2}m{v}_p^2$

And

$U_p=\frac{G{M}_{E}m}{r_p}$

Substitute the value of $K_a$, $K_p$,role="math" localid="1655736829112" $U_a$and$U_P$in the equation (iii).

Substitute the value of $V_p$in the above equation.

Substitute the value of $r_p$and $r_a$in the above equation.

Substitute $6.673×{10}^{−11}{\text{ N³¾}}^{\text{2}}/{\text{kg}}^{\text{2}}$for $G$, $6.37×{10}^6\text{″¾}$for $R_E$, $\text{400}×{\text{10}}^3\text{″¾}$for $h_p$, $\text{4000}×{\text{10}}^3\text{″¾}$for $h_a$, and $5.97×{10}^{24}\text{ k²µ}$ for $M_E$in the above equation.

And

$\begin{array}{l}\frac{v_p}{v_a}=1.53\\ v_p=1.53v_a\end{array}$

Substitute the value of $v_a$in the above equation.

$\begin{array}{c}{v}_p=1.53×5.52×{10}^3\text{″¾}/\text{s}\\ =8.4×{10}^3\text{″¾}/\text{s}\end{array}$

Hence the speed of apogee and speed of perigee are$5.52×{10}^3\text{″¾}/\text{s}$ and$8.4×{10}^3\text{″¾}/\text{s}$ .

d)

The spacecraft escape from earths then the total energy is expressed as,

$E=K+U$

Here$K$ and$U$are the kinetic and gravitational potential energy. For determine speed,$E=0$. Then the total energy expressed as,

$K+U=0$ ...(iv)

At perigee, kinetic and gravitational potential energy is expressed as,

$K_p=\frac{1}{2}m{v}_p^2$

And

$U_p=−\frac{G{M}_{E}m}{r_p}$

Substitute the value of $r_p$,in the above equation.

$U_p=−\frac{G{M}_{E}m}{h_p+R_E}$

Substitute the value of $K_p$, and $U_p$in the equation (iv).

$\begin{array}{c}\frac{1}{2}m{v}_p^2−\frac{G{M}_{E}m}{h_p+R_E}=0\\ v_p=\sqrt{\frac{2G{M}_E}{h_p+R_E}}\end{array}$

Substitute $6.673×{10}^{−11}{\text{ N³¾}}^{\text{2}}/{\text{kg}}^{\text{2}}$for $G$, $6.37×{10}^6\text{″¾}$for $R_E$, $\text{400}×{\text{10}}^3\text{″¾}$for $h_P$and $5.97×{10}^{24}\text{ k²µ}$for $M_E$in the above equation.

$\begin{array}{c}{v}_p=\sqrt{\frac{2×6.673×{10}^{−11}{\text{ N³¾}}^{\text{2}}/{\text{kg}}^{\text{2}}×5.97×{10}^{24}\text{ k²µ}}{\text{400}×{\text{10}}^3\text{″¾}+6.37×{10}^6\text{″¾}}}\\ =1.085×{10}^4\text{″¾}/\text{s}\end{array}$

Hence the increase of speed of is,$(1.085×{10}^4\text{″¾}/\text{s}−8.4×{10}^3\text{″¾}/\text{s})=2.4×{10}^3\text{″¾}/\text{s}$

At apogee, kinetic and gravitational potential energy is expressed as,

$K_a=\frac{1}{2}m{v}_a^2$

And

$U_a=−\frac{G{M}_{E}m}{r_a}$

Substitute the value of $r_a$,in the above equation.

$U_a=−\frac{G{M}_{E}m}{h_a+R_E}$

Substitute the value of $K_a$, and$U_a$ in the equation (iv).

Substitute$6.673×{10}^{−11}{\text{ N³¾}}^{\text{2}}/{\text{kg}}^{\text{2}}$ for $G$,$6.37×{10}^6\text{″¾}$ for$R_E$ ,$\text{4000}×{\text{10}}^3\text{″¾}$ for$h_a$ and$5.97×{10}^{24}\text{ k²µ}$ for$M_E$ in the above equation.

$\begin{array}{c}{v}_p=\sqrt{\frac{2×6.673×{10}^{−11}{\text{ N³¾}}^{\text{2}}/{\text{kg}}^{\text{2}}×5.97×{10}^{24}\text{ k²µ}}{\text{4000}×{\text{10}}^3\text{″¾}+6.37×{10}^6\text{″¾}}}\\ =8.76×{10}^3\text{″¾}/\text{s}\end{array}$

Hence the increase of speed of is,$(8.76×{10}^3\text{″¾}/\text{s}−5.52×{10}^3\text{″¾}/\text{s})=3.24×{10}^3\text{″¾}/\text{s}$