beth12345 4 years ago What would be the gravitational field strength 2000km above the Earth? 50km above the Earth?

1. anonymous

Gravity decreases with altitude, since greater altitude means greater distance from the Earth's center. so the gravity at 2000 km above the earth surface will be very less as compare to 50 km above the earth surface.

2. anonymous

$F=GmM/(R _{e}+r)^2$ where M is mass of the Earth & Re is radius of the Earth

3. beth12345

what is Gm and r?

4. anonymous

r is distance of mass to Earth and G is gravitation constant that equal to 6.63*10^34(i'm not sure about G's value but i think that.many of physics book has this eg holliday)

5. anonymous

G is the gravitational constant and m is the mass of any object on the earth surface it is not r it is h which means

6. anonymous

i think hosein it is h because the question stats that is about the altitude

7. beth12345

uhm, I don't really know the numbers to plug into the equation

8. anonymous

yea here r is h

9. anonymous

you can't solve this,bath?

10. anonymous

so u can put 2000km as h or 5o as h but u cannot solve it

11. beth12345

ya

12. beth12345

i don't know what m is

13. anonymous

m is mass of 2nd body

14. anonymous

m is the mass of a any object on the earth surface or the object above the earth surface e.g a satellite

15. beth12345

do you know what value I would plug into it?

16. anonymous

mass of an object e.g 40 kg but is negligible as compared to earths mass

17. anonymous

G=6.67*10(-11) N(m/kg)^2 M = 5.9722 × 1024 kg

18. anonymous

subtitute the values given if u have any word problem

19. anonymous

you ought be consider that in formula

20. beth12345

ok so would the equation then become $G= 6.67\times10^{-11}$ $M= 5.98 \times10^{24}$ $R e = 6.38 \times10^{6 }$ $h= (2000)or (50)$

21. anonymous

yea put it in formula

22. anonymous

you should these standard values in the formula

23. beth12345

and then $((6.67\times10^{-11}) \times 40) \div ((6.38 \times 10^{6}) + 2000)$

24. anonymous

where is Earth mass? and denominator has power 2

25. anonymous

but accelaration due to gravity too comes in the formula

26. anonymous

we forgot the main thing

27. anonymous

what thing?sgholap100

28. anonymous

accelaration due to gravity

29. anonymous

ican't understand your goal?can you explane

30. anonymous

you can use gRe^2 instead of GMe

31. anonymous

as the question say that the object is above earth surface accelaration due to gravity(g=9.8m/s^2) too comes then the formula becomes

32. anonymous

your goal is?$mg=GMm/R _{e}^2$ am i right?

33. beth12345

$((6.67 \times 10 ^{-11}) \times 40) \times 5.98 \times 10 ^{24}) \div(6.38 \times 10 ^{6} +2000)^{2}$

34. anonymous

oh yeah i am sorry its was my mistake i was considering Effect of altitude on gravity @hosein u r right

35. anonymous

ithink it's right.

36. anonymous

yes you r right

37. beth12345

so would the answer be 391.7 ?

38. beth12345

ok ya I think I got it, thanks guys :)

39. anonymous

welcome but do one thing neglect the m and h in the formula as they are very small as compared to M and R OK ^_^