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irakobra [83]
3 years ago
6

A space probe is launched from Earth headed for deep space. At a distance of 10,000 miles from Earth's center, the gravitational

force on it is 600lb, what is the size of the force when it is at each of the following distances from the Earths surface?
a. 20,000 miles
b. 30,000 miles
c. 100,000 miles
Physics
2 answers:
just olya [345]3 years ago
8 0

gravitational force varies based on 1/r^2

when you're double the distance =10,000 to 20,000, the force is 4 times smaller so on and so forth.

As force is proportional to 1 / {distance squared}, the force will be 1 / 2^2 (i.e. 1/4) of the force at the reference distance (i.e. 1/4 * 600 = 150 lb)



Pani-rosa [81]3 years ago
3 0
Grav Force = GMm/r squared. M is earth's mass. m is the probe. r is distance to centre of earth.Given that 600 = GMm/(10,000^2).GMm = 600x(10,000^2) for 10, 000 milesGMm = ? x (20,000^2) for 20, 000 milestherefore600x(10,000^2)=? x (20,000^2)therefore [600x(10,000^2)]/(20,000^2)=? [600x(1x1)]/(2x2)=? [600/4]=? ?=150lb.Same procedure for b and c.Please ask if you want further help.
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3 years ago
A 62-kg person jumps from a window to a fire net 20.0 m directly below, which stretches the net 1.4 m. Assume that the net behav
gayaneshka [121]

Answer:

a) x = 0.098

b) x = 2.72 m

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(a) To find the stretch of the fire net when the same person is lying in it, you can assume that the net is like a spring with constant spring k. It is necessary to find k.

When the person is falling down he acquires a kinetic energy K, this energy is equal to the elastic potential energy of the net when it is max stretched.

Then, you have:

K=U\\\\\frac{1}{2}mv^2=\frac{1}{2}kx^2        (1)

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v: velocity of the person just when he touches the fire net = ?

x: elongation of the fire net = 1.4 m

Before the calculation of the spring constant, you calculate the final velocity of the person by using the following formula:

v^2=v_o^2+2gy

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y: height from the person jumps = 20.0m

v=\sqrt{2gy}=\sqrt{2(9.8m/s^2)(20.0m)}=14\frac{m}{s}

With this value you can find the spring constant k from the equation (1):

mv^2=kx^2\\\\k=\frac{mv^2}{x^2}=\frac{(62kg)(14m/s)^2}{(1.4m)^2}=6200\frac{N}{m}

When the person is lying on the fire net the weight of the person is equal to the elastic force of the fire net:

W=F_e\\\\mg=kx

you solve the last expression for x:

x=\frac{mg}{k}=\frac{(62kg)(9.8m/s^2)}{6200N/m}=0.098m

When the person is lying on the fire net the elongation of the fire net is 0.098m

b) To find how much would the net stretch, If the person jumps from 38 m, you first calculate the final velocity of the person again:

v=\sqrt{2gy}=\sqrt{2(9.8m/s^2)(38m)}=27.29\frac{m}{s}

Next, you calculate x from the equation (1):

x=\sqrt{\frac{mv^2}{k}}=\sqrt{\frac{(62kg)(27.29m/s)^2}{6200N/m}}\\\\x=2.72m

The net fire is stretched 2.72 m

5 0
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