document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=114345>Question 114345</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A NASA satellite makes a circular orbit around a planet of unknown diamter at an altitude of one mile above the planet.<BR>\n" );
document.write( "Then NASA boosted the satellite to an altitude one mile higher(an altitude of two miles above planet.) In the second orbit, the satellite travels how many miles farther than it travelled on the first orbit?<BR>\n" );
document.write( "(C=2pir)\r<BR>\n" );
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document.write( ">>>>>>>>>>>>PLEASE I NEED YOUR HELP.....THANKS IN ADVANCE </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #83208 by </B><A HREF=/tutors/aboutme.mpl?userid=solver91311><B>solver91311(24713)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=solver91311><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/cgi-bin/show-face.mpl?user=solver91311><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=83208\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> You don't know the diameter of the planet, so let's call that <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=D%5Bp%5D ALIGN=MIDDLE  ALT=\"D%5Bp%5D\"   DISABLED_style= \"max-width:90%; height:auto;\" > miles.  The first orbit is one mile above the planet, so we know that the diameter of the orbit is <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=D%5Bp%5D%2B2 ALIGN=MIDDLE  ALT=\"D%5Bp%5D%2B2\"   DISABLED_style= \"max-width:90%; height:auto;\" >.  We had to add 2 because 1 mile of altitude adds 1 mile to the radius, and the diameter is twice that.  Now, if we move out to a point 2 miles above the surface, the diameter of the new orbit is <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=D%5Bp%5D%2B4 ALIGN=MIDDLE  ALT=\"D%5Bp%5D%2B4\"   DISABLED_style= \"max-width:90%; height:auto;\" >.\r<BR>\n" );
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document.write( "We know that the distance travelled in the first orbit is the circumference of the circle with diameter <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=D%5Bp%5D%2B2 ALIGN=MIDDLE  ALT=\"D%5Bp%5D%2B2\"   DISABLED_style= \"max-width:90%; height:auto;\" >, or <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=C%5B1%5D=%28D%5Bp%5D%2B2%29%2Api ALIGN=MIDDLE  ALT=\"C%5B1%5D=%28D%5Bp%5D%2B2%29%2Api\"   DISABLED_style= \"max-width:90%; height:auto;\" >, and the distance travelled in the second orbit is  <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=C%5B2%5D=%28D%5Bp%5D%2B4%29%2Api ALIGN=MIDDLE  ALT=\"C%5B2%5D=%28D%5Bp%5D%2B4%29%2Api\"   DISABLED_style= \"max-width:90%; height:auto;\" >.\r<BR>\n" );
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document.write( "Now we can subtract:  <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=C%5B2%5D-C%5B1%5D ALIGN=MIDDLE  ALT=\"C%5B2%5D-C%5B1%5D\"   DISABLED_style= \"max-width:90%; height:auto;\" > => <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28%28D%5Bp%5D%2B4%29pi%29-%28%28D%5Bp%5D%2B2%29pi%29 ALIGN=MIDDLE  ALT=\"%28%28D%5Bp%5D%2B4%29pi%29-%28%28D%5Bp%5D%2B2%29pi%29\"   DISABLED_style= \"max-width:90%; height:auto;\" > => <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28D%5Bp%5Dpi%2B4pi%29-%28D%5Bp%5Dpi%2B2pi%29 ALIGN=MIDDLE  ALT=\"%28D%5Bp%5Dpi%2B4pi%29-%28D%5Bp%5Dpi%2B2pi%29\"   DISABLED_style= \"max-width:90%; height:auto;\" > => <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=2pi ALIGN=MIDDLE  ALT=\"2pi\"   DISABLED_style= \"max-width:90%; height:auto;\" > miles.  Not too far, really, just a little over six and a quarter miles.\r<BR>\n" );
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document.write( "Hope this helps,<BR>\n" );
document.write( "John </SPAN>\n" );
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