When a satellite is revolving into the orbit around a planet then we can say
net centripetal force on the satellite is due to gravitational attraction force of the planet, so we will have
![F_g = F_c](https://tex.z-dn.net/?f=F_g%20%3D%20F_c)
![\frac{GM_pM_s}{r^2} = \frac{M_s v^2}{r}](https://tex.z-dn.net/?f=%5Cfrac%7BGM_pM_s%7D%7Br%5E2%7D%20%3D%20%5Cfrac%7BM_s%20v%5E2%7D%7Br%7D)
now we can say that kinetic energy of satellite is given as
![KE = \frac{1}{2}M_s v^2](https://tex.z-dn.net/?f=KE%20%3D%20%5Cfrac%7B1%7D%7B2%7DM_s%20v%5E2)
![KE = \frac{GM_sM_p}{2r}](https://tex.z-dn.net/?f=KE%20%3D%20%5Cfrac%7BGM_sM_p%7D%7B2r%7D)
also we know that since satellite is in gravitational field of the planet so here it must have some gravitational potential energy in it
so we will have
![U = -\frac{GM_sM_p}{r}](https://tex.z-dn.net/?f=U%20%3D%20-%5Cfrac%7BGM_sM_p%7D%7Br%7D)
so we can say that energy from the fuel is converted into kinetic energy and gravitational potential energy of the satellite
Answer:
I would say all of the above.
Explanation:
Look below for more examples
Answer:
41.4* 10^4 N.m^2/C
Explanation:
given:
E= 4.6 * 10^4 N/C
electric field is 4.6 * 10^4 N/C and square sheet is perpendicular to electric field so, area of vector is parallel to electric field
then electric flux = ∫ E*n dA
= ∫ 4.6 * 10^4 * 3*3
= 41.4* 10^4 N.m^2/C
2e min :)) pls park braliest