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sertanlavr [38]
2 years ago
9

NASA decides to put a 128 kg satellite into orbit over the planet Venus because they want to take pictures. The satellite is 37,

000,000 m above the surface of venus. 
Calculate the gravitational field at that altitude. m/s2
how do you calculate this
Physics
1 answer:
ycow [4]2 years ago
6 0
1).  Look up the acceleration of gravity on the surface of Venus.  It's 8.87 m/s².
(That's about 10% less than on Earth.)

2).  Look up the radius of Venus.  It's 6051.8 km.  
(That's about 5% less than Earth's.)

3).  Remember that the gravitational field (acceleration, force) changes
opposite to the square of the distance from the planet ... or the distance
between any two masses that are gravitating.

Now we have enough information to do the calculating.  Notice that the question
only asks for the planet's "gravitational field" (acceleration) way out there.  That
has nothing to do with the satellite's mass, or whatever you decide to put out there,
or even if there's nothing out there at all.  It's just a characteristic of Venus at that
distance from it.

The distances we need to compare are the distances from the center of Venus.

On the surface (distance from the center is the radius of Venus), it's 6,051,800 m.
In the orbit, it's 37,000,000 m.

The ratio is (37,000,000 / 6,051,800) = about 6.1 .

The gravity way out in the orbit is less than on the surface, and by the
same amount as the square of the distance ratio.

Surface gravity = 8.87

Gravity out at the orbit = 8.87 divided by (the distance ratio)²

Gravity = 8.87 / (6.1)²

Better way:  Gravity = 8.87 x (6,051,800/37,000,000)² = <em>0.2373 m/sec²</em>
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Starting from a pillar, you run 200 m east (the + x-direction) at an average speed of 5.0 m/s and then run 280 m west at an aver
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Explanation:

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Suppose the battery in a clock wears out after moving thousand coulombs of charge through the clock at a rate of 0.5 Ma how long
Ksivusya [100]

Answer:

Hello your question is poorly written below is the complete question

Suppose the battery in a clock wears out after moving Ten thousand coulombs of charge through the clock at a rate of 0.5 Ma how long did the clock run on does battery and how many electrons per second slowed?

answer :

a) 231.48 days

b) n = 3.125 * 10^15

Explanation:

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current rate = 0.5 mA

<u>A) Determine how long the clock run on the battery. use the relation below</u>

q = i * t ----- ( 1 )

q = charge , i = current , t = time

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hence  t = 2 * 10^7 secs

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<u>B) Determine how many electrons per second flowed </u>

q = n*e ------ ( 2 )

n = number of electrons

e = 1.6 * 10^-19

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back to equation 2

n ( number of electrons ) = q / e = ( 0.5 * 10^-3 ) / ( 1.6 * 10^-19 )

hence : n = 3.125 * 10^15

8 0
2 years ago
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