## beth12345 Group Title What would be the gravitational field strength 2000km above the Earth? 50km above the Earth? 2 years ago 2 years ago

1. sgholap100 Group Title

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. hosein Group Title

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

3. beth12345 Group Title

what is Gm and r?

4. hosein Group Title

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. sgholap100 Group Title

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. sgholap100 Group Title

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

7. beth12345 Group Title

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

8. hosein Group Title

yea here r is h

9. hosein Group Title

you can't solve this,bath?

10. sgholap100 Group Title

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

11. beth12345 Group Title

ya

12. beth12345 Group Title

i don't know what m is

13. hosein Group Title

m is mass of 2nd body

14. sgholap100 Group Title

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 Group Title

do you know what value I would plug into it?

16. sgholap100 Group Title

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

17. hosein Group Title

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

18. sgholap100 Group Title

subtitute the values given if u have any word problem

19. hosein Group Title

you ought be consider that in formula

20. beth12345 Group Title

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. hosein Group Title

yea put it in formula

22. sgholap100 Group Title

you should these standard values in the formula

23. beth12345 Group Title

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

24. hosein Group Title

where is Earth mass? and denominator has power 2

25. sgholap100 Group Title

but accelaration due to gravity too comes in the formula

26. sgholap100 Group Title

we forgot the main thing

27. hosein Group Title

what thing?sgholap100

28. sgholap100 Group Title

accelaration due to gravity

29. hosein Group Title

ican't understand your goal?can you explane

30. hosein Group Title

you can use gRe^2 instead of GMe

31. sgholap100 Group Title

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. hosein Group Title

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

33. beth12345 Group Title

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

34. sgholap100 Group Title

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

35. hosein Group Title

ithink it's right.

36. sgholap100 Group Title

yes you r right

37. beth12345 Group Title

so would the answer be 391.7 ?

38. beth12345 Group Title

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

39. sgholap100 Group Title

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 ^_^