sykik
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If anyone's wondering Part Bi was to calculate the energy need to move probe from Earth to the Position X. Can Anyone explain this please? I looked at the Markscheme and I don't understand what it's trying to say. (It was something about Earth-sun and Earth- Moon distance). Thanks
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atsruser
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(Original post by sykik)
If anyone's wondering Part Bi was to calculate the energy need to move probe from Earth to the Position X. Can Anyone explain this please? I looked at the Markscheme and I don't understand what it's trying to say. (It was something about Earth-sun and Earth- Moon distance). Thanks
The distance from the Sun to the Earth is huge, compared with the distance from the Sun to the Moon. So to a very good approximation, the work done in moving from the Earth to the Moon due to the Sun's gravitational field is 0, when compared to the work done due to the Earth's gravitational field.

To put it another way, we can consider the Sun's potential field to be constant between Earth and the Moon, but that's not true for the Earth's potential field.
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sykik
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(Original post by atsruser)
The distance from the Sun to the Earth is huge, compared with the distance from the Sun to the Moon. So to a very good approximation, the work done in moving from the Earth to the Moon due to the Sun's gravitational field is 0, when compared to the work done due to the Earth's gravitational field.

To put it another way, we can consider the Sun's potential field to be constant between Earth and the Moon, but that's not true for the Earth's potential field.
Yh That's what exactly Markscheme said but I don't understand that even if the distance is huge why would it neglect sun's gravitation field? Basically can you explain the bit in bold. Thanks
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atsruser
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(Original post by sykik)
Yh That's what exactly Markscheme said but I don't understand that even if the distance is huge why would it neglect sun's gravitation field? Basically can you explain the bit in bold. Thanks
Calculate the difference in the sun's potential between a point on the earth's surface and a point on the moon using U=-\frac{GM_s}{r} and you will find it isn't very large.

Since the work done is the change in potential between 2 points, you won't do much work against the sun's gravitational field moving from the earth to the moon.
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samb1234
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(Original post by sykik)
Yh That's what exactly Markscheme said but I don't understand that even if the distance is huge why would it neglect sun's gravitation field? Basically can you explain the bit in bold. Thanks
G is inversely proportional to r^2. Therefore over a large distance g is very small compared to g at closer distances

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sykik
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(Original post by atsruser)
Calculate the difference in the sun's potential between a point on the earth's surface and a point on the moon using U=-\frac{GM_s}{r} and you will find it isn't very large.

Since the work done is the change in potential between 2 points, you won't do much work against the sun's gravitational field moving from the earth to the moon.

(Original post by samb1234)
G is inversely proportional to r^2. Therefore over a large distance g is very small compared to g at closer distances

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Ahhhh Thanks A lot!!
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