Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

Q: 8.24 A spaceship is stationed on Mars. How much energy must be expended on the spaceship to launch it out of the solar system? Mass of the space ship $\small =1000\hspace{1mm}kg$ ; mass of the sun $\small =2 \times 10^3^0\hspace{1mm}kg$ ; mass of mars $\small =6.4 \times 10^2^3\hspace{1mm}kg$ ; radius of mars $\small =3395\hspace{1mm}km$ ;radius of the orbit of mars $\small =2.28 \times 10^8\hspace{1mm}km$ ; $\small G=6.67 \times 10^-^1^1\hspace{1mm}Nm^2kg^-^2$

Answers (1)

best_answer

Firstly, the potential energy of spaceship due to the sun is given by :

$=\ \frac{-GMm_s}{r}$

and the potential energy of spaceship due to mars is given by :

$=\ \frac{-GMm_m}{R}$

It is given that the spaceship is stationary so its kinetic energy is zero.

Thus the total energy of spaceship is :

$=\ \frac{-GMm_s}{r}\ +\ \frac{-GMm_m}{R}$

Thus the energy needed to launch the spaceship is :

$=\ \frac{GMm_s}{r}\ +\ \frac{GMm_m}{R}$

or $=\ 6.67\times 10^{-11}\times 10^3 \left ( \frac{2\times 10^{30}}{2.28\times 10^{11}}\ +\ \frac{6.4\times 10^{23}}{3.395\times 10^6} \right )$

or $=\ 596.97\times 10^9\ J$

or $=\ 6\times 10^{11}\ J$

Posted by

Devendra Khairwa

View full answer

Crack CUET with india's "Best Teachers"

HD Video Lectures
Unlimited Mock Tests
Faculty Support

cuet_ads

Similar Questions

Trending Articles/News

anam-khan

CBSE Board Exams 2025: Guidelines issued for Class 10, 12 practicals

anam-khan

Board Exam Date Sheet 2025 Live: ICSE, ISC, CBSE, MP time table out; BSEB routine soon

anam-khan

Board exam dates 2025: CBSE, CISCE, UP, MP, Maharashtra, TN time table out; state-wise updates

Latest Question