Question

An astronaut takes a ball of mass $m$ from earth to space. He throws the ball into a circular orbit about earth at an altitude of 318.5 km. From earth's surface to the orbit, the change in total mechanical energy of the ball is $x \frac{GM_em}{21R_e}$. The value of $x$ is
(take $R_e = 6370 \, \text{km}$):

• A. 11
• B. 9
• C. 12
• D. 10

Answer 1

Answer: (A)

11

Answer 2

Step 1: Data and formula for mechanical energy The total mechanical energy at the Earth’s surface is:

$T.E_i = -\frac{GM_e m}{R_e},$

where:

• $R_e = 6370 \, \text{km}$ (radius of Earth),
• $G$ is the gravitational constant,
• $M_e$ is the mass of Earth.

The altitude of the orbit is given as $h = 318.5 \, \text{km}$. Approximate:

$h \approx \frac{R_e}{20}.$

The total mechanical energy in the orbit is:

$T.E_f = -\frac{GM_e m}{2(R_e + h)}.$

Step 2: Substitute $h \approx \frac{R_e}{20}$

$T.E_f = -\frac{GM_e m}{2 \left( R_e + \frac{R_e}{20} \right)} = -\frac{GM_e m}{2 \left( \frac{21R_e}{20} \right)}.$

Simplify:

$T.E_f = -\frac{10GM_e m}{21R_e}.$

Step 3: Change in mechanical energy The change in total mechanical energy is:

$\Delta E = T.E_f - T.E_i.$

Substitute:

$\Delta E = \left( -\frac{10GM_e m}{21R_e} \right) - \left( -\frac{GM_e m}{R_e} \right).$

Simplify:

$\Delta E = -\frac{10GM_e m}{21R_e} + \frac{21GM_e m}{21R_e}.$

$\Delta E = \frac{11GM_e m}{21R_e}.$

Thus, $x = 11$.

Final Answer: $x = 11$.