Question

On February 7, 1999, NASA launched a spacecraft with the ambitious mission of making a close encounter with a comet, collecting samples from its tail, and returning the samples to Earth for analysis. This spacecraft, appropriately named Stardust, took almost five years to rendezvous with its objective-comet Wild 2 (pronounced "Vilt 2")-and another two years to return its samples. The reason for the long round trip is that the spacecraft had to make three orbits around the Sun, and also an Earth Gravity Assist (EGA) flyby, to increase its speed enough to put it in an orbit appropriate for the encounter.
When Stardust finally reached comet Wild 2 on January 2, 2004, it flew within 147 miles of the comet's nucleus, snapping pictures and collecting tiny specks of dust in the glistening coma. The approach speed between the spacecraft and the comet at the encounter was a relatively "slow" 6200 m/s, so that dust particles could be collected safely without destroying the vehicle. Note that "slow" is put in quotation marks; after all, 6200 m/sis still about six times the speed of a rifle bullet!
The roughly spherical comet Wild 2 has a radius of 2.7 km, and the acceleration due to gravity on its surface is 0.00010 g . The two curves in the figure(Figure 1)show the surface acceleration as a function of radius for a spherical comet with two different masses, one of which corresponds to comet Wild 2. Also indicated are radii at which these two hypothetical comets have densities equal to that of ice and granite.
The Stardust spacecraft is still in space; only its small return capsule came back to Earth. It has now been given a new assignment-to visit and photograph comet Tempel 1, the object of the Deep Impact collision on July 4, 2005. This mission, called New Exploration of Tempel 1 (NExT), is scheduled to make its close encounter with comet Tempel 1 on February 14, 2011.
Suppose comet Wild 2 had a small satellite in orbit around it, just as Dactyl orbits asteroid 243 Ida (see page 390).

0.00025 Density of granite 0.0002 0.00015 Acceleration of gravity Density of ice 0.0001 Density of granite 0.00005 1.5 Density of ice 2 2.5 Radius of comet (km) 3.5

8.2h, 0.93h, 5.8h, 2.9h

question: If this satellite were to orbit at twice the radius of the comet, what would be its period of revolution?

Answer 1

Acceleration due to gravity on the surface of planet = g^'= 0.00010 g

GMm R2 = mg

M = mass of comet, m = mass of any object on its surface

GM = R² g'

Now, By Kepler's Law,

r = orbital radius, R = Comet radius

$T^{2^$ =

4114,3/GM

=

4114,3/0 R

=

4114 (2R)/R

=

321T-R g

=   

32 * 3.14 * 3.14 * 2.7 * 103 0.00019

=

8.683*10*

T =
V8.683 * 108
=
2.946 + 104
seconds