The energy required to take a satellite to a height ‘h’ above Earth surface (radius of Earth =6.4×10^{3}_{ }km) is E_{1 }and kinetic energy required for the satellite to be in a circular orbit at this height is E_{2}. Then value of h for which E_{1} and E_{2 }are equal, is

a. |
1.28×10 |
b. |
6.4×10 |

c. |
3.2×10 |
d. |
1.6×10 |

The energy required to take a satellite to a height ‘h’ above Earth surface (radius of Earth =6.4×10^{3}_{ }km) is E_{1 }and kinetic energy required for the satellite to be in a circular orbit at this height is E_{2}. Then value of h for which E_{1} and E_{2 }are equal, is

a. |
1.28×10 |
b. |
6.4×10 |

c. |
3.2×10 |
d. |
1.6×10 |

1 Answer

127 votes

U_{surface }+E_{1 }=U_{h}

KE of satellite is zero at earth surface & at height h

E_{1}=

E_{1}=G

E_{1}=

Gravitational attraction F_{G} = ma_{c} =

E_{2}

mv^{2}=

E_{2} =

E_{1} =E_{2}

= 3200km

127 votes

127