document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=4070>Question 4070</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Please help I am not sure how to solve the following!\r<BR>\n" );
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document.write( "s=1/2 gt^2  (solve for g) </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #1794 by </B><A HREF=/tutors/aboutme.mpl?userid=khwang><B>khwang(438)</B></A><img src=\"/images/stars/stars-10.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=khwang><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=1794\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let the altitude be h and the radius of the earch be R.<BR>\n" );
document.write( "Also, assume the velociety the of the satellite is V<BR>\n" );
document.write( " the central force mV^2/(R+h) = GmM/(R+h)^2<BR>\n" );
document.write( " or V^2 = GM/(R+h) <BR>\n" );
document.write( " so V = sqrt(GM/(R+h))...(1)<BR>\n" );
document.write( " (where M is the mass of the earth , m is the mass of the satellite<BR>\n" );
document.write( "  G is the gravitational constant))<BR>\n" );
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document.write( " The length of the orbit of the satellite is <BR>\n" );
document.write( " 2pi(R+h) = V* 1.5 ( 90 min = 1.5 hrs)<BR>\n" );
document.write( " Solve 2pi(R+h) = 1.5 sqrt(GM/(R+h))...(2)<BR>\n" );
document.write( " for h (by taking squares on both sides) then you can get the answer <BR>\n" );
document.write( " of h(altitude)\r<BR>\n" );
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document.write( " Let T be the time for a satellite to orbit (of radius R+h) the earth.<BR>\n" );
document.write( " Use (1), we have<BR>\n" );
document.write( " T = 2pi (R+h)/ V =  2pi (R+h) /sqrt(GM/(R+h)) =  2pi(R+h)^(3/2)/sqrt(GM)<BR>\n" );
document.write( " so T is an increasing function of h.\r<BR>\n" );
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document.write( " When h = 0, T has min. value 2pi(R)^(3/2)/sqrt(GM).<BR>\n" );
document.write( " But since mg = mGM/R^2 = m V^2/R. (g =9.8 meter/sec^2)<BR>\n" );
document.write( " The corresponding V = sqrt(Rg).<BR>\n" );
document.write( " So, T has min value = 2pi R/ V = 2pi R/ sqrt(Rg) = 2pi sqrt(R/g) \r<BR>\n" );
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document.write( "   <BR>\n" );
document.write( " Of course, you have to find the values of R,M & G by yourself to solve<BR>\n" );
document.write( " (2) and find min T.\r<BR>\n" );
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document.write( " Kenny<BR>\n" );
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