CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

A satellite orbits the earth at a height of 400 km above the surface. How much energy must be expended to rocket the satellite out of the earth's gravitational influence? Mass of the satellite $$= 200\ kg$$; mass of the earth $$= 6.0 \times 10^{24}\ kg$$; radius of the earth $$=6.4 \times 10^6\, m$$; G = $$6.67 \times 10^{-11}\ Nm^2kg^{-2}$$.


Solution

Mass of the Earth, $$M=6.0 \times 10^{24}\ kg$$
$$m=200\ kg$$
 $$R_e=6.4 \times 10^6\ m$$
$$G=6.67 \times 10^{-11}\ Nm^2kg^{-2}$$
Height of the satellite,$$h=400\ km = 4 \times 10^5\ m$$

Total energy of the satellite at height h$$ = (1/2)mv^2+\left(-G \dfrac{M_em}{(R_e+h)}\right)$$
Orbital velocity of the satellite, $$v=\sqrt{\dfrac{GM_e}{R_e+h}}$$

Total energy at height h $$=\dfrac{1}{2}\dfrac{GM_em}{R_e+h}-\dfrac{GM_em}{R_e+h}$$

$$Total\ Energy=-\dfrac{1}{2}\dfrac{GM_em}{R_e+h}$$

The negative sign indicates that the satellite is bound to the Earth. 

Energy required to send the satellite out of its orbit = – (Bound energy)
$$=\dfrac{GM_em}{2(R_e+h)}$$

$$=\dfrac{6.67 \times 10^{-11} \times 6 \times 10^{24} \times 200}{2(6.4 \times 10^6+4 \times 10^5)}$$

$$=5.9 \times 10^9\ J$$

If the satellite just escapes from the gravitational field, then total energy of the satellite is zero. Therefore, we have to supply $$5.9 \times 10^9J$$ of energy to just escape it.

Physics
NCERT
Standard XI

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
Same exercise questions
View More


similar_icon
People also searched for
View More



footer-image