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3 years ago

Why is it difficult to drop a probe like Galileo? How did engineers solve this problem?

Physics

1 answer:

icang [17]3 years ago

4 0

Explanation:

Galileo spacecraft was sent to orbit around and study Jupiter. It was launched on October 18, 1989 and entered into Jupiter's orbit on December 7, 1995. It had a lot of firsts in its name and most important of them was the first probe that entered in Jovian atmosphere. The spacecraft carried a probe named <em>Galileo</em> which separated from the spacecraft 5 months before it was to reach Jupiter. The separation was done in such a way that the rendezvous of probe and spacecraft with Jupiter coincided.

The mission was very challenging. Some of them are listed below:

1. Because of Jupiter's immense gravity the speed of the probe reached up to 47 KM/Sec as it approached Jupiter, but to enter into Jupiter's atmosphere it had to be brought down to subsonic level (0.34 KM/Sec). Due to this it had to withstand 228G deceleration.

2. Also, the time of probe entry had to be synchronized with that of the rendezvous of spacecraft with Jupiter. Very precise calculations were needed to achieve this.

3. Jupiter has a very thick atmosphere which would result in very high pressure and temperature for anything entering into it atmosphere. A very strong heat shield was needed to withstand such high pressure and temperature. Out of the 339 kg of the probe 152 kg was just the weight of the heat shield. Around 80 kg of the heat shield was lost during the entry. Its instruments were so sturdy that they worked till a pressure of 23 atm and temperature of 153°C.

4. The deceleration of the probe was further aided by using a parachute during the entry.

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The 10-kg uniform rod is pinned at end

Anton [14]

Supposing that the spring is un stretched when θ = 0, and has a toughness of k = 60 N/m.It seems that the spring has a roller support on the left end. This would make the spring force direction always to the left
Sum moments about the pivot to zero.
10.0(9.81)[(2sinθ)/2] + 50 - 60(2sinθ)[2cosθ] = 0 98.1sinθ + 50 - (120)2sinθcosθ = 0 98.1sinθ + 50 - (120)sin(2θ) = 0
by iterative answer we discover that
θ ≈ 0.465 radians
θ ≈ 26.6º

3 0

3 years ago

The rate traveled from Amarillo to Austin by a bus averages 60 miles per hour. The bus arrived in Austin after eight hours of tr

jeyben [28]

Answer:

the car would arrive after 10 hr to Austin.

Explanation:

bus average from amarllo to austin = 60 miles per hour

time takne by bus to reach austin = 8 hr

automobile average from amarllo to austin = 60 miles per hour

from the information given in the question:

60 mph --- 8 h

48 mph --- x h

By using The inverse variation:

60 : 48 = x : 8

60* 8 = 48* x

480 = 48*x

x = 480/48

x = 10 h

the car would arrive after 10 hr to Austin.

3 0

3 years ago

Suppose a capacitor is fully charged by a battery and then disconnected from the battery. The positive plate has a charge +q and

dimaraw [331]

Answer:-q

Explanation:

Given

Capacitor is charged to a battery and capacitor acquired a charge of q i.e.

+q on Positive Plate and -q on negative Plate.

If the plate area is doubled and the plate separation is reduced to half its initial separation then capacitor becomes four times of initial value because capacitor is given by

$c=\epsilon_0 \cdot \frac{A}{d}$

where A=area of capacitor plate

d=Separation between plates

This change in capacitance changes the Potential such that new charge on the negative plate will remain same -q

8 0

3 years ago

How is Coulomb’s law similar to newton’s law of gravitational force? How is it different

natulia [17]

The similarities and the differences between gravitational and electric force are listed below

Explanation:

- The magnitude of the gravitational force between two objects is given by Newton's law of gravitation:

$F=G\frac{m_1 m_2}{r^2}$

where

$G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2}$ is the gravitational constant

$m_1, m_2$ are the masses of the two objects

r is the separation between them

- Coloumb's law gives instead the strength of the electrostatic force between two charged objects, which is

$F=k\frac{q_1 q_2}{r^2}$

where:

$k=8.99\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

$q_1, q_2$ are the two charges

r is the separation between the two charges

By comparing the two equations, we find the following similarities:

Both the forces are inversely proportional to the square of the distance between the two objects, $F\propto \frac{1}{r^2}$
Both the forces are proportional to the product between the "main quantity" of each force, which is the mass for the gravitational force ( $F\propto m_1 m_2$ ) and the charge for the electric force ( $F\propto q_1 q_2$

Instead, we have the following differences:

The gravitational force is always attractive, since the sign of $m$ is always positive, while the electric force can be either attractive or repulsive, since the sign of $q$ can be either positive or negative
The value of the gravitational costant G is much smaller than the value of the Coulomb's constant, so the gravitational force is much weaker than the electric force